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A constant step
A STORY
The same step every time
Deo climbs the school staircase, one step at a time. Nila waits at the bottom, one hand on the
first step.
“Every step is the same height,” Nila says. “I measured one last week: $15$ cm.”
Deo stops. “Then tell me how high my foot is. It is on the fourth step.”
“Four steps of $15$ cm,” Nila says. “$60$ cm above the floor.”
“And the twentieth step, at the top?”
“$20$ times $15$ is $300$ cm, three metres up. The same step every time makes it easy.”
“Now a harder one,” Deo says. “Climb one step and come down. Then two steps and come down. Then
three, and so on, up to all twenty. How many steps do you climb in all?”
“One, two, three, and on to twenty,” Nila says slowly. “That is a long sum to add by hand.”
“There is a short way,” Deo says.
You will learn the rule for any one step in a pattern like this, and the rule for adding up all
of them.
₹200 goes into a savings box in the first month. Every month after that, ₹40 more goes in than the month before. The monthly deposits climb $200, 240, 280, 320, \cdots$. Your turn: check the gap between each deposit and the one before it. (Answer: every gap is exactly ₹40.)
Let us name that fixed step. A sequence built this way is an ARITHMETIC PROGRESSION, or AP. Each term differs from the one before it by the same fixed number. The fixed number is called $d$, the common difference.
Two questions follow at once. Picture one particular term, far down the list: could you name it without writing out every term before it? Could you add up a whole run of terms just as fast? We answer both questions from the start, not by handing over a ready-made shortcut.
A SCHOLAR INDIA REMEMBERS
Aryabhata wrote a book called the Aryabhatiya in 499 CE, when he was $23$. He lived at Kusumapura, near today’s Patna. His book gave general methods: one method for a whole class of problems. We want the same here. Take the list $3$, $7$, $11$, $15$. We will not reach the 100th number by writing all $100$. We will find one rule that gives any number in the list.
Aryabhata · the hoopoe
Each piece is one term, each piece after the first is four longer, and the marks below give the running total.
Check yourself
Which description correctly defines an arithmetic progression (AP)?
the step from one term to the next stays fixed
each term is double the one before it
each term increases, but by a different amount each time
the terms have no fixed pattern between them
Check your answer
✓ the step from one term to the next stays fixed — (A) An AP is exactly a list where the step from one term to the next stays the same fixed number every time.
each term is double the one before it — Doubling each term is a geometric progression’s rule, not an AP’s — an AP adds the same number, it does not multiply.
each term increases, but by a different amount each time — A changing step size is exactly what an AP rules out — the whole point of $d$ is that it stays the same term after term.
the terms have no fixed pattern between them — An AP is a very specific pattern — a random list with no fixed step would not be an AP at all.
In an AP, what does the common difference $d$ actually control?
how many terms the AP has
the value of the first term
the added amount, term to term
which position a term sits at in the list
Check your answer
how many terms the AP has — The number of terms is $n$, a separate count — $d$ says nothing about how many terms exist.
the value of the first term — The first term is $a$, fixed once at the start — $d$ only governs the step after it.
✓ the added amount, term to term — (C) $d$ is the fixed amount added to a term to produce the next one — that is its whole job in the definition.
which position a term sits at in the list — A term’s position in the list is its index, not $d$ — $d$ is the same number at every position.
An auto fare starts at ₹25 for the first kilometre, then rises by ₹8 for every kilometre after that. In this fare pattern, which quantity plays the role of $d$?
the ₹8 rise per extra kilometre
the ₹25 starting fare
the total number of kilometres travelled
the total fare paid at the end of the ride
Check your answer
✓ the ₹8 rise per extra kilometre — (A) The role of $d$ is always the fixed amount added per step — here, the ₹8 added for each kilometre after the first.
the ₹25 starting fare — ₹25 is the starting value, $a$ — it is fixed once and does not repeat as a step.
the total number of kilometres travelled — The number of kilometres travelled is $n$, how many terms are in play — not the amount added between them.
the total fare paid at the end of the ride — The total fare is an outcome of the pattern, not the fixed step that builds it — that step is the ₹8 rise.
You have met sequences before. Now name the fixed step between two terms. Try each check below. It takes a minute.
The common difference (Class 9, Ganita Manjari, page 181). Here: $d$ is the fixed step this progression adds, term after term. Check: $1, 5, 9, 13, \dots$ has common difference $4$.
An arithmetic progression (Class 9, Ganita Manjari, page 180). Here: a list is tested by checking that every gap has the same value. Check: $3, 6, 9, 12$ has equal gaps of $3$, so it is an AP.
The nth term formula (Class 9, Ganita Manjari, page 182). Here: $a_n = a + (n - 1) d$ finds the 30th term without listing 29 others. Check: $t_n = 1 + (n - 1) \cdot 4$ gives $t_3 = 1 + 2 \cdot 4 = 9$.
An explicit rule for a sequence (Class 9, Ganita Manjari, page 176). Here: this one rule generates every term, without writing each one out. Check: $u_n = 2n - 1$ gives $u_3 = 2 \cdot 3 - 1 = 5$.
A sequence’s recursive rule (Class 9, Ganita Manjari, page 183). Here: each term here is built from the one before it, deriving the nth term. Check: $t_1 = 1$, $t_2 = t_1 + 4 = 5$.
Sum of the first n numbers (Class 9, Ganita Manjari, page 184). Here: adding forwards and backwards this way derives the sum formula. Check: $1 + 2 + 3 + 4 = 10$, and $4 \cdot 5 / 2 = 10$.
If any of these felt new, read the page named before going on.
The COMMON DIFFERENCE of an AP is the fixed number $d$ added to each term to get the next one. Pick two consecutive terms, $a_k$ and $a_{k + 1}$. You can find $d$ yourself: $d = a_{k + 1} - a_k$, the later term minus the earlier one, in that order.
*$d$ can be positive, negative, or zero: all three are legal common differences.* A salary rising by a fixed annual raise has a positive $d$. A savings amount shrinking by a fixed instalment each month has a negative $d$. A list where every term repeats the same number has $d = 0$. We still call it an AP: just a flat one, a fixed step of zero.
Your turn: find $d$ for the AP $6, 3, 0, -3, \cdots$. (Answer: $d = 3 - 6 = -3$, so each term is three less than the one before it.) Getting the subtraction order backwards, using $a_k - a_{k + 1}$ instead, flips the sign of $d$ and every later prediction.
The same four terms, subtracted both ways round: only the order the key term names gives d.
A SCHOLAR INDIA REMEMBERS
Is $2$, $4$, $8$, $16$ an AP? Count the steps before you answer. $4 - 2 = 2$. $8 - 4 = 4$. $16 - 8 = 8$. The steps are $2$, $4$ and $8$. They are not equal. So this list is not an AP. An AP takes the same step every time.
Aryabhata · the hoopoe
Check yourself
How is the common difference $d$ of an AP correctly found from two consecutive terms?
$d = a_k - a_{k + 1}$, the earlier term minus the later one
$d = a_{k + 1} - a_k$, the later term minus the earlier one
$d = a_{k + 1} + a_k$, the two terms added together
$d = a_{k + 1} / a_k$, the later term divided by the earlier one
Check your answer
$d = a_k - a_{k + 1}$, the earlier term minus the later one — Reversing the order flips the sign of $d$ — for a rising AP this wrongly makes $d$ negative.
✓ $d = a_{k + 1} - a_k$, the later term minus the earlier one — (B) $d$ is defined as the later consecutive term minus the earlier one, in that fixed order.
$d = a_{k + 1} + a_k$, the two terms added together — Adding two terms gives their sum, not the step between them — $d$ comes only from subtraction.
$d = a_{k + 1} / a_k$, the later term divided by the earlier one — A divided ratio between terms is a geometric progression’s common ratio — an AP’s $d$ comes from subtraction, not division.
An AP’s terms get smaller with every step. What must be true of its common difference $d$?
$d$ must be positive
$d$ must be zero
$d$ must be negative
$d$ could be any sign, since the terms are shrinking either way
Check your answer
$d$ must be positive — A positive $d$ makes each term bigger than the last — the opposite of the falling pattern described here.
$d$ must be zero — A zero $d$ keeps every term identical — it would never fall, only stay the same.
✓ $d$ must be negative — (C) Terms that keep shrinking are exactly what a negative $d$ produces — each step subtracts rather than adds.
$d$ could be any sign, since the terms are shrinking either way — The sign of $d$ is precisely what decides whether the AP rises or falls — it cannot be either sign here.
A water tank holds 40 litres at 6 a.m. and loses 5 litres every hour after that. Reading the litres left each hour as an AP, what is $d$?
$5$
$-5$
$40$
$8$
Check your answer
$5$ — The tank is losing water, not gaining it — the step is $-5$, not $5$.
✓ $-5$ — (B) Each hour subtracts 5 litres from the term before it, so the common difference is $-5$.
$40$ — 40 litres is the starting amount, $a$ — it is not the amount lost each hour.
$8$ — No 8-litre quantity appears in this scenario at all — the only rate given is 5 litres lost per hour.
The GENERAL FORM of an AP has first term $a$ and common difference $d$. Let us write it out: $a, a + d, a + 2 d, a + 3 d, \cdots$. Every term after the first is built the same way: you take the one before it and add $d$.
The first term is often written $a_1$, or simply $a$, when no confusion is possible. You can see the pattern continue: the second term is $a_2$, the third $a_3$, and so on, one subscript higher for every step along the list.
$a$ and $d$ between them fix everything. Nothing else changes what the list looks like. Your turn: write out the first four terms of the AP with $a = 5$ and $d = 3$. (Answer: $5, 8, 11, 14$.)
Check yourself
Which is the correct general form of an AP with first term $a$ and common difference $d$?
$a, 2 a, 3 a, 4 a, \cdots$
$a, a + d, a + 4 d, a + 9 d, \cdots$
$a, a + d, a + 2 d, a + 3 d, \cdots$
$a, d, 2 d, 3 d, \cdots$
Check your answer
$a, 2 a, 3 a, 4 a, \cdots$ — Multiplying $a$ by whole numbers is not how an AP is built at all — every step must add the fixed $d$.
$a, a + d, a + 4 d, a + 9 d, \cdots$ — Squaring the term count breaks the fixed-step rule — the gap between consecutive terms would keep growing, not stay at $d$.
✓ $a, a + d, a + 2 d, a + 3 d, \cdots$ — (C) Each term adds one more $d$ than the last, starting from $a$ with zero additions.
$a, d, 2 d, 3 d, \cdots$ — Every term of an AP must still include the starting value $a$ — dropping it after the first term breaks the pattern.
In the general form, the third term is written $a + 2 d$. Why $2 d$ and not $3 d$?
because $d$ is always counted starting from the second term, not the first
because the third term is always exactly two-thirds of the way through the AP
reaching the third term needs only two additions of $d$ from the first term
because $a$ itself already counts as one addition of $d$
Check your answer
because $d$ is always counted starting from the second term, not the first — There is no separate counting rule for $d$ — every step is counted the same way, from the first term onward.
because the third term is always exactly two-thirds of the way through the AP — An AP’s terms are not measured by a fraction of the way through — the count of additions is what matters.
✓ reaching the third term needs only two additions of $d$ from the first term — (C) Counting from the first term, only two steps of $d$ are needed to arrive at the third term — the first term itself needs none.
because $a$ itself already counts as one addition of $d$ — $a$ is the starting point before any step is taken — it is not itself one of the additions being counted.
The first term of an AP is often written $a_1$ instead of plain $a$. Why does this notation help?
it shows that the first term is somehow different from every other term in the AP
it labels every term, including the first, by its position number
it means the first term must always equal $1$
Check your answer
it shows that the first term is somehow different from every other term in the AP — The first term follows the exact same pattern as every later one — the subscript is only a position label, not a sign of difference.
✓ it labels every term, including the first, by its position number — (B) Writing $a_1, a_2, a_3, \cdots$ gives every term, including the first, a position label that matches the rest of the list.
it means the first term must always equal $1$ — The subscript $1$ marks position, not value — the first term can be any number, positive, negative, or zero.
A list $a_1, a_2, a_3, \cdots$ is an AP when the difference between every pair of consecutive terms is the same. Compute $a_2 - a_1$, then $a_3 - a_2$, and so on. If every difference you compute comes out the same, the list is an AP, and that shared value is its $d$.
One matching pair is never enough. You should check at least two or three consecutive differences before deciding. We can be fooled by a single coincidence: two terms may differ by the right amount purely by chance.
Your turn: test $2, 4, 8, 16, \cdots$ the same way. (Answer: $4 - 2 = 2$, but $8 - 4 = 4$, so the differences disagree and this list is not an AP, even though the pattern looked tempting.)
Once a list has actually been confirmed as an AP, the shortcut runs the other way. A single pair of consecutive terms is now enough to read off $d$: every other pair matches it too.
Points on a straight line have equal gaps between them, and points that curve away do not.
Check yourself
What must be true for a list of numbers to count as an AP?
the difference between the first and last terms is a whole number
every term is bigger than the one before it
every consecutive pair shares the same difference
the terms can be arranged in any order to look evenly spaced
Check your answer
the difference between the first and last terms is a whole number — Checking only the first and last terms says nothing about what happens between them — every consecutive pair must match.
every term is bigger than the one before it — Many rising lists are not APs — squares and cubes rise too, without ever sharing one fixed step.
✓ every consecutive pair shares the same difference — (C) A list is an AP exactly when every consecutive pair shares one common difference — no exceptions anywhere in the list.
the terms can be arranged in any order to look evenly spaced — An AP’s terms come in a fixed order — rearranging a list to look spaced out is not the same as it already being an AP.
To confirm that a list of 5 numbers is an AP, how many consecutive differences must be checked?
just the first 1 difference
5 differences, one for every term
only the difference between the first and last term
all 4 consecutive differences
Check your answer
just the first 1 difference — One matching pair says nothing about the other three — a list can match for one step and break later.
5 differences, one for every term — Differences come from PAIRS of consecutive terms, not from single terms — 5 terms give only 4 such pairs.
only the difference between the first and last term — The overall first-to-last gap can look consistent even when the steps in between are not — every internal pair still needs checking.
✓ all 4 consecutive differences — (D) A list of 5 terms has 4 consecutive pairs, and every one of them must share the same difference before the list counts as an AP.
Is $4, 9, 14, 19$ an AP? If so, what is $d$?
yes, an AP with $d = 5$
yes, an AP with $d = 4$
no, because the terms are not all even numbers
no, because the list has only 4 terms
Check your answer
✓ yes, an AP with $d = 5$ — (A) Every consecutive difference here is $5$ ($9-4$, $14-9$, $19-14$), so the list is an AP with $d = 5$.
yes, an AP with $d = 4$ — $4$ is the first term, $a$ — the actual step between consecutive terms here works out to $5$, not $4$.
no, because the terms are not all even numbers — An AP’s terms can be any numbers at all — evenness plays no role in the definition, only a fixed common difference does.
no, because the list has only 4 terms — How many terms a list has does not decide whether it is an AP — only whether every consecutive difference matches does.
A student claims: ‘$1, 4, 9, 16$ is an AP, because each term is bigger than the one before it.’ What is wrong with this claim?
nothing is wrong — any steadily rising list is an AP
the consecutive differences here are $3, 5, 7$, not equal
the claim is wrong because the list should start from $0$, not $1$
the claim is wrong because the list has too few terms to tell
Check your answer
nothing is wrong — any steadily rising list is an AP — A steadily rising list is not automatically an AP — the differences must all be the SAME number, and here they are not.
✓ the consecutive differences here are $3, 5, 7$, not equal — (B) The differences between consecutive terms are $3$, $5$, and $7$ — three different values, so the list fails the AP test outright.
the claim is wrong because the list should start from $0$, not $1$ — An AP can start from any number at all — the starting value is never the issue here; the unequal differences are.
the claim is wrong because the list has too few terms to tell — The four given terms already show unequal differences — no further terms are needed to rule this list out.
An AP can end, or it can run forever. A FINITE AP has a fixed number of terms and stops at a LAST TERM, usually written $l$. An INFINITE AP never stops: after any term, another one always follows, by the same rule as every term before it.
Let us look at the AP $147, 148, 149, \cdots, 157$. Is it finite or infinite? You can tell from the last term alone: it is finite, because the list is written with an explicit last term, $157$, and it stops there. Your turn: is $1, 4, 7, 10, \cdots$ finite or infinite? (Answer: infinite, since it is written with no last term and the pattern continues without limit.)
The number of seats in one row of a hall is a finite AP: the row has an end. The multiples of 3, listed in order, form an infinite AP, since there is no largest multiple of 3.
The top row stops at a marked last chip, and the bottom row keeps going, trailing off with three dots.
Check yourself
What distinguishes a finite AP from an infinite one?
a finite AP has a common difference $d$; an infinite AP does not
a finite AP stops at a fixed last term; an infinite AP never stops
a finite AP always has a negative $d$; an infinite AP always has a positive $d$
a finite AP has only whole-number terms; an infinite AP can have any terms
Check your answer
a finite AP has a common difference $d$; an infinite AP does not — Both finite and infinite APs have a fixed common difference — what differs is only whether the list ever stops.
✓ a finite AP stops at a fixed last term; an infinite AP never stops — (B) A finite AP has a definite last term, often written $l$; an infinite AP keeps producing another term after every one.
a finite AP always has a negative $d$; an infinite AP always has a positive $d$ — The sign of $d$ says whether an AP rises or falls — it has no bearing on whether the list is finite or infinite.
a finite AP has only whole-number terms; an infinite AP can have any terms — Both finite and infinite APs can hold any kind of number as a term — finiteness is only about whether the list ends.
Which of these is an example of a finite AP?
the natural numbers $1, 2, 3, 4, \cdots$, counted without end
the odd numbers $1, 3, 5, 7, \cdots$, listed forever
any AP where $d$ is a whole number
hourly temperatures recorded through one 24-hour day
Check your answer
the natural numbers $1, 2, 3, 4, \cdots$, counted without end — Counting the natural numbers never stops at a fixed last term — that is exactly the infinite case, not the finite one.
the odd numbers $1, 3, 5, 7, \cdots$, listed forever — Listing the odd numbers ‘forever’ has no last term at all — it is infinite, not finite.
any AP where $d$ is a whole number — Whether $d$ is a whole number has nothing to do with the list stopping — only a fixed last term makes an AP finite.
✓ hourly temperatures recorded through one 24-hour day — (D) A single day gives exactly 24 hourly readings and then stops — a fixed last term, so the list is finite.
A stack of 15 logs has the shortest log on top; each log below is 4 cm longer than the one above it. Reading the log lengths from top to bottom as an AP, which term is $l$?
the length of the bottom, 15th log
the length of the top, shortest log
the 4 cm difference between one log and the next
Check your answer
✓ the length of the bottom, 15th log — (A) $l$ names the last term of a finite AP — here, the length of the bottommost, 15th log, where the stack ends.
the length of the top, shortest log — The top log’s length is the first term, $a$ — $l$ specifically names the LAST term, at the bottom of the stack.
the 4 cm difference between one log and the next — The 4 cm gap is the common difference, $d$ — $l$ refers to a specific term’s length, not the step between terms.
The $n$th term of an AP with first term $a$ and common difference $d$ is *$a_n = a + (n - 1) d$*. Reaching the $n$th term means starting at $a$ and taking $(n - 1)$ steps of size $d$. That is one step fewer than the term number, because the first term itself needs no step at all. The full build-up of this formula, term by term, follows shortly.
For a finite AP with last term $l$, the same formula gives $l = a + (n - 1) d$, where $n$ is the total number of terms. The last term is simply the $n$th term, under a different name.
We now have four quantities linked together: $a$, $d$, $n$, and $a_n$. Know any three of them, and you can find the fourth.
Your turn: first term $5$, common difference $2$, what is the 6th term? (Answer: $a_6 = 5 + (6 - 1) \cdot 2 = 15$.)
Each dot is one nth-term value, climbing in a straight line when the common difference is positive and falling when it is negative.
A SCHOLAR INDIA REMEMBERS
Will your rule work for the next case too? Test $a_n = a + (n - 1) d$ where you can check it. For $3$, $7$, $11$, $15$, we have $a = 3$ and $d = 4$. The 4th term is $3 + 3 \times 4 = 15$. That matches the list. Now use it for the 100th term: $3 + 99 \times 4 = 399$.
Aryabhata · the hoopoe
Check yourself
Which is the correct formula for the $n$th term of an AP with first term $a$ and common difference $d$?
$a_n = a + (n - 1) d$
$a_n = a + n d$
$a_n = n a + d$
$a_n = a + (n + 1) d$
Check your answer
✓ $a_n = a + (n - 1) d$ — (A) Reaching the $n$th term needs exactly $(n - 1)$ additions of $d$ from the first term — one fewer step than the term’s own number.
$a_n = a + n d$ — This adds one $d$ too many — the first term $a_1$ needs zero additions, not one, so the count must be $(n-1)$.
$a_n = n a + d$ — $a$ is never multiplied in this formula — only $d$ is scaled by the step count, and $a$ stays as it is.
$a_n = a + (n + 1) d$ — Adding $(n+1)$ steps overshoots by two beyond the correct $(n-1)$ — the term after $a_n$, not $a_n$ itself, would need that many.
Why does the $n$th-term formula use $(n - 1)$ instead of $n$?
the first term needs no addition at all
because $d$ is defined as one less than the actual step size
because the formula only works when $n$ is even
because the last term is always excluded from the count
Check your answer
✓ the first term needs no addition at all — (A) The first term is $a$ itself, with zero additions — every later term adds one $d$ per step, so the count of additions is always one less than the term number.
because $d$ is defined as one less than the actual step size — $d$ is exactly the step size, nothing less — the $(n-1)$ comes from counting steps, not from redefining $d$.
because the formula only works when $n$ is even — The formula holds for every positive whole number $n$, odd or even — there is no such restriction.
because the last term is always excluded from the count — This formula finds any term $a_n$, not specifically a last term — nothing here excludes any term from being counted.
In $a_n = a + (n - 1) d$, why must $n$ be a positive whole number?
because $d$ would become negative otherwise
because the formula only applies to APs with a positive first term
$n$ names a term’s position in the list
because otherwise the sum $S_n$ could not be calculated
Check your answer
because $d$ would become negative otherwise — The sign of $d$ depends only on whether the AP rises or falls — it has nothing to do with what values $n$ may take.
because the formula only applies to APs with a positive first term — $a$ can be positive, negative, or zero — the restriction here is on $n$, a position count, not on the first term’s value.
✓ $n$ names a term’s position in the list — (C) $n$ names WHICH term in the list is meant — first, second, third, and so on — and a position in a list is always a positive whole number.
because otherwise the sum $S_n$ could not be calculated — This question is about the term formula $a_n$, not the sum $S_n$ — the reason $n$ stays positive is about position, not about a different formula.
Seats in a hall are numbered starting at 101, increasing by 1 along each row. Reading the seat numbers of one row as an AP, what is the 20th seat number in that row?
$121$
$2020$
$120$
$119$
Check your answer
$121$ — Reaching the 20th seat needs only $19$ steps of $1$ from the first seat, not $20$ — that extra step overshoots by one.
$2020$ — Seat numbers are not multiplied together — each seat adds one more step of $d$ to the one before it.
Let us take an AP with first term $a$ and common difference $d$. The sum of its first $n$ terms is *$S_n = n/2 \cdot [2 a + (n - 1) d]$*. When the last term $l$ is known instead of $d$, the same sum can be written $S_n = n/2 \cdot (a + l)$. That is the average of the first and last term, multiplied by how many terms there are. The full derivation, using a trick from Gauss, follows shortly.
Two equivalent forms exist, for two different situations. Use the $d$-form when you know the common difference. Use the $l$-form when you know the last term instead. Finding $S_n$ needs $n$, and either $d$ or $l$: never mix the two forms inside one calculation.
Your turn: first term $4$, last term $20$, $10$ terms. Find $S_{10}$. (Answer: $S_{10} = 10/2 \cdot (4 + 20) = 120$.)
A SCHOLAR INDIA REMEMBERS
Before you use the formula, guess the size of the answer. $1 + 2 + 3 + \dots + 100$ has $100$ terms. On average a term is about $50$. So the sum is about $100 \times 50 = 5000$. The formula gives $100/2 \times (1 + 100) = 5050$. The guess was close, so the answer makes sense.
Aryabhata · the hoopoe
Check yourself
Which is the correct formula for the sum $S_n$ of the first $n$ terms of an AP?
$S_n = n \cdot [a + (n - 1) d]$
$S_n = n/2 \cdot [2 a + n d]$
$S_n=n/2 \cdot[2a+(n-1)d]$
$S_n = a + (n - 1) d$
Check your answer
$S_n = n \cdot [a + (n - 1) d]$ — Without the $n/2$, every pair the derivation forms gets double-counted — the halving is what corrects for that.
$S_n = n/2 \cdot [2 a + n d]$ — The bracket needs $(n-1)$ steps of $d$, matching the last term’s own formula — using $n$ overshoots by one step of $d$.
✓ $S_n=n/2 \cdot[2a+(n-1)d]$ — (C) This formula adds all $n$ terms at once, using the average of the first and last term multiplied by how many terms there are.
$S_n = a + (n - 1) d$ — This expression is the formula for a single term, $a_n$ — the sum of all the terms needs the full $S_n$ formula, not this one.
In the sum formula, why does the factor $n/2$ appear at all?
because an AP always has an even number of terms
because $a$ and $l$ are always exactly half of each other
because only half the terms in an AP are actually counted
forward-plus-backward double-counts every term
Check your answer
because an AP always has an even number of terms — An AP can have any number of terms, odd or even — the $n/2$ comes from the doubling in the derivation, not from a rule about term count.
because $a$ and $l$ are always exactly half of each other — $a$ and $l$ can be any two numbers at all — nothing requires one to be half the other.
because only half the terms in an AP are actually counted — Every one of the $n$ terms is included in the sum — none are skipped; the halving corrects for double-counting, not for missing terms.
✓ forward-plus-backward double-counts every term — (D) Adding the forward and backward versions of the sum gives $n$ pairs that each total $2a + (n-1)d$ — since that counts every term twice over, dividing by $2$ gives the actual sum.
When is $S_n = n/2 \cdot (a + l)$ the more useful sum formula to reach for?
when the AP has a negative common difference
when $l$ is already known, skipping $d$
when the AP is infinite
when $a$ is unknown
Check your answer
when the AP has a negative common difference — Both sum formulas work for any sign of $d$ — which one is more convenient depends on what is already known, not on $d$’s sign.
✓ when $l$ is already known, skipping $d$ — (B) This form needs only the first term, the last term, and the term count — it skips finding $d$ altogether when $l$ is already known.
when the AP is infinite — An infinite AP never reaches a last term, so $l$ does not exist for it — this formula needs a finite AP with a known $l$.
when $a$ is unknown — $S_n = n/2 \cdot (a + l)$ still needs $a$ — what it removes is the need to compute $d$ first, when $l$ is already known.
A company hires 6 employees in its first year, and 4 more each year after that. Reading the yearly hires as an AP, what is the total number hired over the first 10 years?
$46$
$260$
$240$
$400$
Check your answer
$46$ — $46$ is the single figure for the 10th year alone, $a_{10}$ — the question asks for the total across all 10 years, which is $S_{10}$.
$260$ — The bracket needs $(n-1) \cdot d$, which is $9 \cdot 4 = 36$, not $10 \cdot 4 = 40$ — using $n$ there overshoots by one step of $d$.
$400$ — Multiplying one year’s figure by 10 assumes every year hired the same number — but hiring grows by 4 each year, so the terms are not all equal.
A term and a sum are linked, but never equal. The $n$th term of an AP equals the difference between two consecutive sums: *$a_n = S_n - S_{n - 1}$*, for $n \geq 2$. A single term is exactly what a sum GAINS by reaching one term further, not the sum itself.
For $n = 1$ there is nothing to subtract: $a_1 = S_1$, since summing just one term gives that term back.
Let us take the AP $3, 7, 11, \cdots$, where $S_3 = 21$ and $S_4 = 36$. Then $a_4 = S_4 - S_3 = 15$. Your turn: check $a_4$ against the ordinary formula. (Answer: $3 + (4 - 1) \cdot 4 = 15$, the same answer, confirming the relation both ways.)
This relation LINKS $a_n$ and $S_n$. It never makes the two equal. The difference between them is exactly what the next section asks you to catch.
Check yourself
For $n \geq 2$, how is a single term $a_n$ correctly recovered from sums $S_n$ and $S_{n-1}$?
$a_n = S_n - S_{n - 1}$
$a_n = S_n + S_{n - 1}$
$a_n = S_{n - 1} - S_n$
$a_n = S_n / n$
Check your answer
✓ $a_n = S_n - S_{n - 1}$ — (A) $S_n$ already includes every term up to $a_n$; subtracting off $S_{n-1}$, which stops one term earlier, leaves exactly $a_n$ on its own.
$a_n = S_n + S_{n - 1}$ — Adding $S_n$ and $S_{n-1}$ counts every term up to $a_{n-1}$ twice — only subtraction isolates the single new term, $a_n$.
$a_n = S_{n - 1} - S_n$ — Reversing the order gives the negative of $a_n$ — $S_n$, the larger running total, must come first.
$a_n = S_n / n$ — Dividing $S_n$ by $n$ gives the AVERAGE of the first $n$ terms, not the specific term $a_n$ added last.
If $S_7 = 84$ and $S_6 = 66$ for an AP, what is $a_7$?
$150$
$12$
$18$
$84$
Check your answer
$150$ — Adding the two sums does not isolate a single term — only subtracting $S_6$ from $S_7$ leaves the 7th term on its own.
$12$ — $84 / 7 = 12$ is the AVERAGE of the first 7 terms — the actual 7th term still needs $S_7 - S_6$.
✓ $18$ — (C) $a_7 = S_7 - S_6 = 84 - 66 = 18$ — the running total through 7 terms minus the running total through 6 leaves just the 7th term.
$84$ — $84$ is the total of all 7 terms added together, $S_7$ — the single 7th term alone is a smaller number, found by subtracting $S_6$.
The relation $a_n = S_n - S_{n - 1}$ needs a special case for $a_1$. Why?
because $a_1$ is always equal to $d$
$S_0$ does not exist — there are no earlier terms
because the formula only works for even-numbered terms
because $S_n$ cannot be computed for small values of $n$
Check your answer
because $a_1$ is always equal to $d$ — $a_1$ and $d$ are independent quantities — the first term can take any value, unrelated to the common difference.
✓ $S_0$ does not exist — there are no earlier terms — (B) $S_{n-1}$ at $n = 1$ would mean $S_0$, the sum of zero terms — there is nothing to subtract, so $a_1 = S_1$ directly instead.
because the formula only works for even-numbered terms — The formula $a_n = S_n - S_{n-1}$ works for every term from $n = 2$ onward, odd or even — the actual exception is only at $n = 1$.
because $S_n$ cannot be computed for small values of $n$ — $S_n$ is well defined for every $n = 1, 2, 3, \cdots$ — the actual gap is only $S_0$, which has no terms to sum at all.
A charity’s running total of donations after 9 days is ₹4500, and after 8 days it is ₹3800. How much was donated on the 9th day alone?
₹8300
₹4500
₹700
₹500
Check your answer
₹8300 — Adding the two totals does not isolate a single day’s donation — subtracting the day-8 total from the day-9 total does.
₹4500 — ₹4500 is the total collected over all 9 days — the 9th day alone is a smaller amount, found by subtracting the day-8 total.
✓ ₹700 — (C) The 9th day’s donation is the day-9 running total minus the day-8 running total: $4500 - 3800 = 700$.
₹500 — ₹4500 / 9 = ₹500 is an AVERAGE across all 9 days — the actual 9th-day donation still comes from subtracting the day-8 total.
Any two terms of an AP can be written using $a_n = a + (n - 1) d$. We now have two linear equations in the two unknowns $a$ and $d$. Write the two equations, then solve them together, and you will find both unknowns. Every other term of the AP then follows from the same formula.
Subtract the two equations first. Doing so removes $a$ immediately, since $a$ appears in both equations with the same coefficient. That leaves one equation in $d$ alone, which you can solve directly. Your turn: in the equations $a + 3 d = 11$ and $a + 8 d = 26$, which unknown cancels when you subtract? (Answer: $a$ cancels, leaving $5 d = 15$.)
The rise of fifteen over the run of five gives the common difference, three, the slope of this line.
Check yourself
Knowing any two terms of an AP gives you what, in terms of finding $a$ and $d$?
the value of $d$ directly, with no further work needed
one equation with two unknowns, which cannot be solved
two linear equations in $a$ and $d$
the sum $S_n$ of the AP directly
Check your answer
the value of $d$ directly, with no further work needed — Two terms do not hand over $d$ for free — they must first be written as two equations and then solved together.
one equation with two unknowns, which cannot be solved — Each known term gives its own equation — two known terms give two equations, which together CAN be solved for $a$ and $d$.
✓ two linear equations in $a$ and $d$ — (C) Each known term plugs into $a_n = a + (n-1)d$ to give one equation — two known terms give two such equations, enough to solve for both $a$ and $d$.
the sum $S_n$ of the AP directly — Two terms let you solve for $a$ and $d$, not for $S_n$ directly — $S_n$ would need its own formula applied afterward.
Given two equations $a + 3 d = 11$ and $a + 8 d = 26$, what is the most direct way to find $d$?
subtract the first equation from the second, eliminating $a$
add the two equations together
guess values of $a$ and $d$ until both equations happen to work
divide the second equation by the first
Check your answer
✓ subtract the first equation from the second, eliminating $a$ — (A) Subtracting removes $a$ from both equations at once, since it appears identically in both, leaving a single equation in $d$ alone.
add the two equations together — Adding does not cancel either unknown here — both $a$ and $d$ remain mixed together, so neither is isolated.
guess values of $a$ and $d$ until both equations happen to work — Guessing can work by luck, but subtracting the equations solves for $d$ directly and reliably, with no trial and error needed.
divide the second equation by the first — Dividing one equation by another does not cancel $a$ cleanly here — subtraction is the operation that removes it.
In an AP, the 3rd term is $8$ and the 7th term is $20$. What are $a$ and $d$?
$a = 8$, $d = 20$
$a = 5$, $d = 3$
$a = 2$, $d = 4$
$a = 2$, $d = 3$
Check your answer
$a = 8$, $d = 20$ — $8$ and $20$ are term VALUES, not $a$ and $d$ themselves — they must be substituted into equations and solved, not read off directly.
$a = 5$, $d = 3$ — $d = 3$ is correct, but $a$ then comes from $8 - 2 \cdot 3 = 2$, not $5$ — check the substitution back into the first equation.
$a = 2$, $d = 4$ — Subtracting the equations gives $4d = 12$, so $d = 3$, not $4$ — recheck the subtraction of the two equations.
✓ $a = 2$, $d = 3$ — (D) $a + 2d = 8$ and $a + 6d = 20$; subtracting gives $4d = 12$, so $d = 3$, and then $a = 8 - 2 \cdot 3 = 2$.
The $r$th term from the end of a finite AP is found by a trick. Let us reverse the AP and read the same list backwards, starting from the last term $l$. You will find it is itself an AP, with first term $l$ and common difference $-d$: every step that added $d$ moving forward now subtracts $d$ moving backward.
Reversing turns a “from the end” question into an ordinary $n$th-term question, just starting from $l$ instead of $a$. *The $r$th term from the end is $l + (r - 1) (-d)$, the ordinary formula applied to the reversed AP.*
Counting from the end restarts the count at $l$. Do not confuse a term’s position counted from the end with its position counted from the start: check which direction a question means before you begin. The two counts run in opposite directions along the same list. Your turn: if an AP has $d = 4$, what is the common difference of its reversed form? (Answer: $-4$.)
Check yourself
What is the correct formula for the $r$th term from the end of a finite AP with last term $l$ and common difference $d$?
$l + (r - 1) d$
$l + r d$
$a + (r - 1) d$
$l + (r-1)(-d)$
Check your answer
$l + (r - 1) d$ — Reversing the AP’s direction flips the sign of the step — counting from the end needs $-d$, not the original $d$.
$l + r d$ — Reaching the $r$th term from the end needs $(r-1)$ steps back from $l$, not $r$ — the last term itself needs zero steps.
$a + (r - 1) d$ — Counting from the end starts at $l$, the last term, not at $a$, the first — this formula answers a different question.
✓ $l + (r-1)(-d)$ — (D) Reversing the AP gives a new AP starting at $l$ with common difference $-d$, so the same $(r-1)$-step rule applies to it.
Why can the ordinary nth-term formula be reused, unchanged in form, to find a term counted from the end of a finite AP?
because $l$ always equals $a$ in a finite AP
because every finite AP secretly has $d = 0$
reversing gives another AP: first term $l$, step $-d$
because counting from the end only works when $n$ is odd
Check your answer
because $l$ always equals $a$ in a finite AP — $a$ and $l$ are the first and last terms of the same AP and are equal only when the AP has a single term — otherwise they differ.
because every finite AP secretly has $d = 0$ — A finite AP can have any common difference at all — most are not constant, so $d = 0$ is not a general fact about them.
✓ reversing gives another AP: first term $l$, step $-d$ — (C) A finite AP read backwards is still an AP — it just starts at $l$ instead of $a$, and steps by $-d$ instead of $d$.
because counting from the end only works when $n$ is odd — The reversal argument works for a finite AP of any length, odd or even — no such restriction exists.
A finite AP begins $50, 44, 38, \cdots$ and ends at the last term $-10$. What is the 6th term from the end?
$14$
$-46$
$-10$
$20$
Check your answer
$14$ — The 6th term from the end needs only 5 steps of 6 from $-10$, not 6 — that extra step overshoots by one.
$-46$ — Counting from the end needs the reversed common difference, $+6$, not the original $-6$ carried over unchanged.
$-10$ — $-10$ is the very last term, the 1st from the end — the question asks for the 6th term counting backward from there.
✓ $20$ — (D) $d = -6$, so reversing gives first term $-10$ and common difference $6$; the 6th term from the end is $-10 + (6 - 1) \cdot 6 = -10 + 30 = 20$.
Here is the mistake most worth catching: asked for a specific term, it is tempting to compute the sum of the first $n$ terms instead. Or the reverse can happen: asked for a running total, you report a single term instead. The two questions sound similar. They are not the same question, and the two answers are almost never the same number.
Let us take the AP $3, 7, 11, 15, 19$. The fifth term is $a_5 = 19$: one single number, the fifth entry on the list and nothing else. The sum of the first five terms is $S_5 = 55$: every one of those five numbers added together. Both numbers describe the same AP. Neither one stands in for the other.
The two are related, but only through *$a_n = S_n - S_{n - 1}$, never through $a_n = S_n$*. A term is what a running total gains by going one step further, not the total itself. Mixing the two is not a small slip. A term and a sum can differ by a wide margin, as $19$ and $55$ already show.
Before you answer a question about an AP, read it again. Ask yourself exactly one thing: does it name a single term, or a total? Your turn: a problem gives the value in the 5th year, and separately asks how much has built up after 5 years. Which formula answers each? (Answer: the value in the 5th year is a term, $a_5$. The total built up after 5 years is a sum, $S_5$.) Get this one read wrong, and the whole calculation goes to the wrong formula.
The single dot on the left is a term, and all the bars on the right added together are a total.
First try
Asked for the 5th term of $3, 7, 11, 15, 19$, I added all five terms and got $a_5 = 55$.
Second look
$a_n$ is one term. $S_n$ is the running total. Here $a_5 = 19$, the fifth term alone, while $S_5 = 55$ is all five added.
A term and a sum answer different questions. Never swap them.
A savings box adds a fixed amount every month. What has it saved by month 9?
Weaker. A reader wants the total saved after 9 months, for a box that holds ₹200 after month 1 and grows by ₹40 every month. They compute $a_9 = 200 + (9 - 1) \cdot 40 = 520$ and give ₹520 as the answer.
But ₹520 is just the amount added in month 9 alone, one single deposit, not everything saved so far. Nine months of deposits, each at least ₹200, must add up to far more than ₹520. The size of that answer alone is the giveaway that the wrong formula was used.
Stronger. The question asks for a total, so it needs $S_n$, not $a_n$. $S_9 = 9/2 \cdot [2 \cdot 200 + (9 - 1) \cdot 40] = 9/2 \cdot [400 + 320] = 9/2 \cdot 720 = 3240$.
The total saved after 9 months is ₹3240, not ₹520. Checking the size of an answer against the question asked catches this slip before it is written down.
Check yourself
A student was asked for the 5th term of the AP $3, 7, 11, 15, 19$ and answered $55$. What went wrong?
nothing is wrong — $55$ is an acceptable way to state the 5th term
the student should have said $15$ instead, the 4th term
$55$ is the SUM of all five terms — the 5th term alone is $19$
Check your answer
nothing is wrong — $55$ is an acceptable way to state the 5th term — A single term and a running sum are never the same number here — $a_5 = 19$ is the 5th term; $55$ is the sum of all five terms.
the student should have said $15$ instead, the 4th term — The 4th term, $15$, is still just a single term — the actual mistake is answering with a SUM, $55$, when a single term was asked for.
✓ $55$ is the SUM of all five terms — the 5th term alone is $19$ — (C) The 5th term is a single value, $a_5 = 19$; $55$ is the running total of all five terms added together, $S_5$, a different quantity entirely.
A student was asked for $S_4$ of the AP $5, 8, 11, 14$ and answered $14$. What went wrong?
nothing is wrong — the last term IS the sum of an AP
$14$ is $a_4$ alone — $S_4 = 5+8+11+14 = 38$
the student should have said $11$ instead, the 3rd term
Check your answer
nothing is wrong — the last term IS the sum of an AP — The last term is only one value in the list — the sum adds up EVERY term, not just the final one.
✓ $14$ is $a_4$ alone — $S_4 = 5+8+11+14 = 38$ — (B) $S_4$ asks for the running total of all four terms added together, $38$; the student instead gave the last single term, $a_4 = 14$.
the student should have said $11$ instead, the 3rd term — $11$ is still just one term among four — the actual mistake is giving a single term at all, when the running total, $S_4$, was asked for.
Asked for $a_4$ of the AP $10, 15, 20, 25, 30$, a student answered $70$. What went wrong?
nothing is wrong — $70$ is a valid way to express $a_4$
$a_4 = 25$ alone — $70$ sums four terms instead
the student should have said $30$ instead, the 5th term
the student should have added all five terms, not just four
Check your answer
nothing is wrong — $70$ is a valid way to express $a_4$ — $a_4$ names one single term, $25$ — a partial sum of several terms answers a different question.
✓ $a_4 = 25$ alone — $70$ sums four terms instead — (B) The 4th term is $25$ by itself; $70 = 10+15+20+25$ is the sum of the first four terms, a different question’s answer.
the student should have said $30$ instead, the 5th term — $30$ is the 5th term, not the 4th — but the deeper error is answering with a sum, $70$, when a single term was asked for.
the student should have added all five terms, not just four — Adding more terms only deepens the term-vs-sum mix-up — the question asks for one single term, not any running total at all.
A ticket counter sells 12 tickets on day 1, and 3 more each following day. ‘How many tickets are sold on day 8?’ asks for which quantity?
$S_8$, the running total sold through day 8
$d$, the daily increase in tickets sold
$n$, the number of days the counter has been open
$a_8$, a single day’s ticket count
Check your answer
$S_8$, the running total sold through day 8 — ‘On day 8’ names one day alone — the total sold through all 8 days is a separate quantity, $S_8$, that this question does not ask for.
$d$, the daily increase in tickets sold — $d = 3$ is the fixed daily INCREASE, not the actual number of tickets sold on day 8 itself.
$n$, the number of days the counter has been open — $n = 8$ names which day is meant, not the answer — the actual ticket count for that day is $a_8$.
✓ $a_8$, a single day’s ticket count — (D) ‘On day 8’ points at one specific day’s figure — that is exactly what $a_8$, a single term, gives.
A library charges a late fee that grows by ₹2 every day a book stays overdue, starting at ₹5 for day 1. ‘What is the TOTAL fee owed after 10 overdue days?’ needs which quantity?
$a_{10}$, the fee charged on day 10 alone
$S_{10}$, the sum of all ten days’ fees
$d$, the daily rise in the fee
$a$, the fee on the very first overdue day
Check your answer
$a_{10}$, the fee charged on day 10 alone — The fee for day 10 alone, $a_{10}$, is just one day’s charge — the question asks for the TOTAL owed across all 10 days, which is $S_{10}$.
✓ $S_{10}$, the sum of all ten days’ fees — (B) ‘Total fee owed after 10 days’ is a running total across every day so far — exactly what $S_n$ gives, not any single day’s fee.
$d$, the daily rise in the fee — $d = 2$ is the fixed daily rise in the fee — it is not itself an amount owed, which is what the question is asking for.
$a$, the fee on the very first overdue day — ₹5 is only the first day’s fee — the total owed after 10 days must add up every day’s fee, not just the first one.
$11 - 5 = 6$ Compute the first consecutive difference.
$17 - 11 = 6$ Now compute a second consecutive difference.
$23 - 17 = 6$ Compute a third difference too, so the pattern is not just a coincidence.
all three differences equal $6$ Every consecutive pair checked gives $a_{k + 1} - a_k = 6$, so the list is an AP.
the AP has $a = 5$ and $d = 6$ Three matching differences, not two, is enough to trust the pattern. The first term and the shared difference read off directly.
$a_4 = 5 + (4 - 1) \cdot 6 = 23$ Check the pattern a different way: the formula gives the 4th term as $23$ too, the same number the list already showed.
Check whether any list of numbers is an arithmetic progression.
Compute the differences Subtract each term from the one after it, at least three times in a row.
Check they all match If every difference comes out the same, the list is an AP, and that shared value is $d$.
Read off a and d The first term of the list is $a$. The common difference is the value you just found.
Check yourself
Which check correctly confirms that $5, 11, 17, 23, \cdots$ is an AP?
$11-5=17-11=23-17=6$
$23 - 5 = 18$, so the common difference is $18$
$11 / 5 = 2.2$, so the common ratio is $2.2$
checking that $5$ and $23$ are both odd numbers
Check your answer
✓ $11-5=17-11=23-17=6$ — (A) Checking every consecutive pair and finding the same value, $6$, each time is exactly what confirms the list is an AP.
$23 - 5 = 18$, so the common difference is $18$ — $18$ is the gap between the FIRST and LAST terms across three steps, not the step between consecutive terms — $d$ comes only from consecutive pairs.
$11 / 5 = 2.2$, so the common ratio is $2.2$ — A common RATIO belongs to a geometric progression’s test — an AP is confirmed by a common DIFFERENCE, found through subtraction.
checking that $5$ and $23$ are both odd numbers — Whether terms are odd or even plays no role in the AP test — only equal consecutive differences do.
For an unfamiliar list of numbers, why must every consecutive pair be checked, rather than just one pair?
because the AP definition only applies to lists longer than 4 terms
because checking more than one pair changes the value of $d$
a coincidence could make one pair match
because the first pair is always the least reliable one to check
Check your answer
because the AP definition only applies to lists longer than 4 terms — The AP definition applies to lists of any length, including very short ones — there is no minimum length requirement.
because checking more than one pair changes the value of $d$ — $d$, if it exists, is a fixed property of the whole list — checking more pairs reveals that value, it does not change it.
✓ a coincidence could make one pair match — (C) A list could share the same difference for one pair by chance and then break the pattern later — checking every pair is what actually confirms an AP.
because the first pair is always the least reliable one to check — No pair is inherently more or less reliable — every consecutive pair must simply match for the list to count as an AP.
Using the same consecutive-difference check, is $12, 19, 26, 32$ an AP?
yes — most of the differences are $7$, which is close enough
yes — the first and last terms differ by $20$, a whole number
no — the differences are $7, 7, 6$, not all equal
Check your answer
yes — most of the differences are $7$, which is close enough — ‘Close enough’ is not the AP test — every single consecutive difference must be exactly equal, with no exceptions.
yes — the first and last terms differ by $20$, a whole number — A whole-number gap between the first and last terms says nothing about whether every step in between matches — only checking every consecutive pair does.
✓ no — the differences are $7, 7, 6$, not all equal — (C) The first two differences are $7$, but the last is only $6$ — since not every consecutive pair matches, the list fails the AP test.
$a_1 = a$ The first term, by definition. Nothing has been added yet.
$a_2 = a_1 + d = a + d$ Each term is the one before it, plus $d$.
$a_3 = a_2 + d = a + 2 d$ Add $d$ once more to the second term.
$a_4 = a_3 + d = a + 3 d$ The same step, repeated.
the coefficient of $d$ is one less than the term number The pattern from four terms already built: 0, 1, 2, 3 additions of $d$ for terms 1, 2, 3, 4.
$a_n = a + (n - 1) d$ *This generalises the pattern: reaching the $n$th term takes $(n - 1)$ additions of $d$, one fewer than the term number.*
$a_4 = a + (4 - 1) d = a + 3 d$ Check the new formula against the pattern already built: putting $n = 4$ in gives $a + 3 d$, matching the fourth term found above.
The steps taken are always one fewer than the term number, which is where the minus one in the formula comes from.
Check yourself
In deriving $a_n = a + (n-1)d$ term by term ($a_2 = a+d$, $a_3 = a+2d$, $a_4 = a+3d$, …), which observation lets the pattern jump straight to $a_n$?
every term after the second is exactly double the one before it
the pattern only holds up to the 4th term shown, and cannot be extended further
the count of additions always trails the term number by exactly one
$d$ gets larger by one unit at every step of the derivation
Check your answer
every term after the second is exactly double the one before it — Each term adds $d$ once more than the last — nothing here doubles; that would be a geometric progression’s pattern, not this one.
the pattern only holds up to the 4th term shown, and cannot be extended further — The derivation’s whole purpose is to show the pattern holds for EVERY term — stopping at the 4th term shown was only for illustration.
✓ the count of additions always trails the term number by exactly one — (C) The worked pattern shows each term’s count of additions trailing its term number by exactly one — spotting that lets the derivation generalise straight to $a_n = a + (n-1)d$.
$d$ gets larger by one unit at every step of the derivation — $d$ stays exactly the same fixed number throughout — what changes from term to term is only how many times it gets added.
Why does the count of additions in the derivation always trail the term number by exactly one?
$a_1$ needs zero additions — counting starts there
because the derivation always skips the second term
because $d$ is only added on even-numbered terms
because the last term in a derivation is always one step short
Check your answer
✓ $a_1$ needs zero additions — counting starts there — (A) $a_1 = a$ needs no addition at all — it is where counting starts — so every later term’s addition count is one behind its own term number.
because the derivation always skips the second term — The derivation moves through every term in order, $a_1, a_2, a_3, \cdots$, with none skipped — the trailing count comes from where counting starts, not from a skip.
because $d$ is only added on even-numbered terms — $d$ is added at EVERY step from one term to the next, odd or even — there is no such parity restriction in the derivation.
because the last term in a derivation is always one step short — This trailing-by-one pattern holds for every term in the derivation, not specifically the last one — it is about where counting starts, not about endings.
A student derived $a_4 = a + 4 d$. What is wrong with this step?
nothing is wrong — $a + 4 d$ correctly gives the 4th term
only 3 additions of $d$ reach the 4th term, not 4
the error is that $a$ should not appear in the formula at all
Check your answer
nothing is wrong — $a + 4 d$ correctly gives the 4th term — Counting the additions carefully, $a_1$ through $a_4$ needs exactly 3 steps of $d$, not 4 — the formula should read $a + 3 d$.
✓ only 3 additions of $d$ reach the 4th term, not 4 — (B) $a_1 = a$, $a_2 = a+d$, $a_3 = a+2d$, $a_4 = a+3d$ — three additions of $d$ get from the first term to the fourth, not four.
the error is that $a$ should not appear in the formula at all — Every term of an AP includes the starting value $a$ — the actual error here is only in how many $d$’s are added, not in whether $a$ belongs.
the salaries form an AP with $a = 15000$ and $d = 800$ The salary starts at ₹15000 and rises by a fixed ₹800 every year, exactly the definition of a common difference.
$a_{12} = 15000 + 8800 = 23800$ Finish the arithmetic.
the salary in the 12th year is ₹23800 *This is one single term of the AP, $a_{12}$, not a running total of everything earned so far.*
$a_{11} = 15000 + (11 - 1) \cdot 800 = 23000$, then $23000 + 800 = 23800$ Check the answer a second way, from the year before it: the 11th year comes to ₹23000, and one more raise of ₹800 gives ₹23800 again.
The chip for year twelve shows one salary, not the total of all twelve years’ raises added together.
Find the salary in a given year, when it starts fixed and rises by a fixed amount every year.
Find a and d The starting salary is $a$. The fixed yearly raise is $d$.
Write the term you need Use $a_n = a + (n - 1) d$ for the year number you want.
Do the arithmetic Multiply $(n - 1)$ by $d$, then add $a$.
Read it as a term, not a total The answer is the salary in that one year, not everything earned up to it.
Check yourself
A job’s monthly salary starts at ₹18000 with an annual increment of ₹1200. Using the same method as the salary example in this chapter, find the salary in the 6th year.
₹24000
₹25200
₹7200
₹108000
Check your answer
✓ ₹24000 — (A) With $a=18000$ and $d=1200$, $a_6 = 18000 + (6-1) \cdot 1200 = 18000+6000=24000$.
₹25200 — Reaching the 6th year needs only 5 annual increments from the starting salary, not 6 — that extra increment overshoots by one.
₹7200 — ₹7200 is only the amount ADDED by year 6 — the actual salary also includes the ₹18000 starting figure, which this option drops.
₹108000 — Salaries here grow by a fixed ADDITION each year, not by multiplication — the starting salary is not scaled up by the year number.
A second job offers the same ₹18000 start but grows by a fixed 5% each year instead of a fixed rupee amount. Why does the AP method from this chapter no longer apply directly to that second job?
because percentages cannot be used in any mathematics chapter
because ₹18000 is too large a starting salary for an AP
a percentage rise is not a fixed rupee step
Check your answer
because percentages cannot be used in any mathematics chapter — Percentages are used all across mathematics — the actual reason the AP method fails here is that the yearly rupee increase is not fixed.
because ₹18000 is too large a starting salary for an AP — An AP’s first term can be any size at all — the issue here is the growth RULE, a percentage, not the size of the starting salary.
✓ a percentage rise is not a fixed rupee step — (C) An AP needs the same rupee amount added every year; a percentage rise adds a different, growing rupee amount each year, so the salaries do not form an AP.
For the chapter’s own salary example ($a=15000$, $d=800$), a student computed the 12th-year salary as $12 \cdot 800 + 15000 = 24600$. What is wrong with this step?
only 11 increments are needed, not 12
nothing is wrong — $24600$ is the correct 12th-year salary
the starting salary ₹15000 should not have been included at all
Check your answer
✓ only 11 increments are needed, not 12 — (A) Reaching the 12th year from the 1st needs $(12-1)=11$ annual increments, not 12 — the correct total is $15000 + 11 \cdot 800 = 23800$.
nothing is wrong — $24600$ is the correct 12th-year salary — Counting the increments carefully, only 11 raises separate the 1st and 12th years, not 12 — the correct salary is ₹23800, not ₹24600.
the starting salary ₹15000 should not have been included at all — The starting salary correctly belongs in every year’s total — the actual error is only in how many increments were added, not in whether the base salary counts.
$a + 3 d = 11$ The 4th term, using $a_n = a + (n - 1) d$ with $n = 4$.
$a + 8 d = 26$ The 9th term, the same formula with $n = 9$.
$5 d = 15$ *Subtracting removes $a$ immediately and leaves one equation in $d$ alone.*
$d = 3$ Divide both sides by 5.
$a = 11 - 3 \cdot 3 = 2$ Substitute $d = 3$ back into the first equation.
the AP is $2, 5, 8, 11, 14, \cdots$ List the terms starting from $a = 2$, adding $d = 3$ each time.
$2 + 3 \cdot 3 = 11$ and $2 + 8 \cdot 3 = 26$ Check both original equations by substituting $a = 2$ and $d = 3$ back in: both come out right, the 4th term $11$ and the 9th term $26$.
Find the first term and the common difference, from any two given terms of an AP.
Write both terms as equations Use $a_n = a + (n - 1) d$ for each given term number.
Subtract the equations This removes $a$ and leaves one equation in $d$ alone.
Solve for d Divide to find $d$.
Substitute back for a Put $d$ into either equation and solve for $a$.
List the AP and check Start at $a$ and add $d$ each time. Check that the two given terms come out right.
Check yourself
In an AP, the 2nd term is $7$ and the 6th term is $23$. Using the same two-equation method as this chapter’s worked example, find $a$ and $d$.
$a = 7$, $d = 23$
$a = 3$, $d = 5$
$a = 4$, $d = 4$
$a = 3$, $d = 4$
Check your answer
$a = 7$, $d = 23$ — $7$ and $23$ are term VALUES, not $a$ and $d$ — the two-equation method must still be solved to find the actual $a$ and $d$.
$a = 3$, $d = 5$ — Subtracting the two equations gives $4d=16$, so $d=4$, not $5$ — recheck the subtraction of the equations.
$a = 4$, $d = 4$ — $d=4$ is correct, but $a$ then comes from $7-4=3$, not $4$ — recheck the substitution back into the first equation.
✓ $a = 3$, $d = 4$ — (D) $a+d=7$ and $a+5d=23$; subtracting gives $4d=16$, so $d=4$, and then $a=7-4=3$.
Which order of steps correctly solves $a+3d=11$ and $a+8d=26$ for $a$ and $d$?
subtract to find $d$ first, then substitute back to find $a$
substitute a guessed value of $a$ into both equations to check which one fits $d$
solve each equation separately for $a$, since $d$ is not needed to find it
Check your answer
✓ subtract to find $d$ first, then substitute back to find $a$ — (A) Subtracting eliminates $a$ cleanly, giving $d$ directly; substituting that $d$ back into either original equation then gives $a$.
substitute a guessed value of $a$ into both equations to check which one fits $d$ — No guessing is needed — subtracting the two equations gives $d$ directly, with no trial and error involved.
solve each equation separately for $a$, since $d$ is not needed to find it — Each equation mixes $a$ and $d$ together — $d$ must be found first, by subtracting the equations, before $a$ can be isolated.
Solving two equations like $a+3d=11$ and $a+8d=26$ together by elimination is the same skill practised in full in which other Class 10 chapter?
Linear Equations in Two Variables
Introduction to Trigonometry
Some Applications of Trigonometry
Check your answer
✓ Linear Equations in Two Variables — (A) Elimination between two linear equations in two unknowns is exactly the method the ‘Pair of Linear Equations in Two Variables’ chapter builds in depth — here it is only reused, not re-taught.
Introduction to Trigonometry — Introduction to Trigonometry covers ratios like sine and cosine — solving two linear equations together is a different chapter’s core skill.
Some Applications of Trigonometry — Some Applications of Trigonometry covers heights and distances — it does not build the two-equation elimination method used here.
the common difference is $d = -3$ $12 - 15 = -3$, confirmed by $9 - 12 = -3$.
the reversed AP has first term $-21$ and common difference $3$ Reversing the AP turns a “from the end” question into an ordinary nth-term question. The reversed list starts from the last term $l = -21$, and $-d$ becomes $3$.
the 8th term from the end is the 8th term of the reversed AP Reversing restarts the count at $l$.
$-21 + (8 - 1) \cdot 3 = -21 + 21 = 0$ Apply the ordinary nth-term formula to the reversed AP, with $r = 8$.
$-21, -18, -15, -12, -9, -6, -3, 0$ List the reversed AP out and count to the 8th term: it lands on $0$ again, the same answer the formula gave above.
The fourth chip from the end and the fifth chip from the front hold the same value in the reversed list.
Find a term counted from the end of a finite AP.
Find d Subtract consecutive terms to get the common difference.
Reverse the AP The reversed list starts at the last term $l$, with common difference $-d$.
Apply the ordinary formula Use $a_n = a + (n - 1) d$ on the reversed AP, with the position counted from the end.
Read off the answer The result is the term you wanted, counted from the end.
Check yourself
A finite AP begins $40, 35, 30, \cdots$ and ends at the last term $-15$. Find the 5th term from the end.
$10$
$5$
$-40$
$-15$
Check your answer
$10$ — The 5th term from the end needs only 4 steps of 5 from $-15$, not 5 — that extra step overshoots by one.
✓ $5$ — (B) $d=-5$, so the reversed AP starts at $-15$ with common difference $5$; the 5th term from the end is $-15+(5-1) \cdot 5=-15+20=5$.
$-40$ — Counting from the end needs the reversed common difference, $+5$, not the original $-5$ carried over unchanged.
$-15$ — $-15$ is the very last term, the 1st from the end — the question asks for the 5th term counting backward from there.
Why does the common difference become $-d$, not $d$, when a finite AP is read from its last term backward?
because the last term $l$ is always negative
because counting from the end always uses subtraction instead of a formula
because $r$ must always be larger than $n$
reversing direction flips the sign of every step
Check your answer
because the last term $l$ is always negative — $l$ can be positive, negative, or zero — the sign flip that matters here is in the STEP, $d$, not in the value of $l$ itself.
because counting from the end always uses subtraction instead of a formula — A formula still applies here — the same nth-term shape, just with $l$ in place of $a$ and $-d$ in place of $d$.
because $r$ must always be larger than $n$ — $r$ and $n$ need no particular relationship to each other here — the negation of $d$ comes from reversing direction, not from comparing $r$ and $n$.
✓ reversing direction flips the sign of every step — (D) The reversed list steps the same size but the opposite way — a step that added $d$ going forward now subtracts $d$ going backward.
For the AP ending at $-21$ with $d=-3$, a student computed the 8th term from the end as $-21+8 \cdot 3=3$. What is wrong?
8 steps were used, but only 7 are needed
nothing is wrong — $3$ is the correct 8th term from the end
the sign of $d$ should not have been reversed at all
Check your answer
✓ 8 steps were used, but only 7 are needed — (A) The 1st term from the end is $l$ itself, needing zero steps — reaching the 8th term from the end needs only $(8-1)=7$ steps, giving $-21+7 \cdot 3=0$, not 8 steps.
nothing is wrong — $3$ is the correct 8th term from the end — Counting carefully, only 7 steps separate the last term from the 8th term counting backward, not 8 — the correct value is $0$, not $3$.
the sign of $d$ should not have been reversed at all — Reversing the sign of $d$ is exactly the correct step here — the actual error is only in the step COUNT, using 8 instead of 7.
Deriving the sum formula, Gauss’s reverse-and-add trick
$S_n = a + (a + d) + (a + 2 d) + \cdots + [a + (n - 1) d]$ Write the sum forwards, term by term.
$S_n = [a + (n - 1) d] + [a + (n - 2) d] + \cdots + (a + d) + a$ Write the identical sum backwards, term by term.
every matched pair totals $2 a + (n - 1) d$ What one line gains going forward, the other loses going backward: every pair totals the same amount.
$2 S_n = n [2 a + (n - 1) d]$ There are $n$ such pairs, one for every term in the sum.
$S_n = n/2 \cdot [2 a + (n - 1) d]$ Divide both sides by 2.
$S_4 = 4/2 \cdot [2 \cdot 5 + (4 - 1) \cdot 6] = 56$ Check the formula against a small case already seen: the AP $5, 11, 17, 23$ has $a = 5$ and $d = 6$. Adding directly, $5 + 11 + 17 + 23 = 56$ too.
The forward row and the backward row pair up so that every one of the five columns adds to 22.
Check yourself
In the Gauss-style derivation, why does every forward-backward pair total the same value, $2a+(n-1)d$?
forward gain always matches backward loss, term for term
because every term in the AP is secretly equal
because $n$ is always an even number in this derivation
because the first and last terms are always equal to each other
Check your answer
✓ forward gain always matches backward loss, term for term — (A) As the forward sum climbs by $d$ at each step, the backward sum falls by exactly the same $d$ — so every paired total stays fixed.
because every term in the AP is secretly equal — The AP’s terms need not be equal at all — the derivation works for any $d$; the pair totals stay fixed because the forward gain always matches the backward loss.
because $n$ is always an even number in this derivation — The derivation holds for any $n$, odd or even — pairing forward and backward sums works regardless of how many terms there are.
because the first and last terms are always equal to each other — $a$ and $l$ are typically different values — what stays fixed across every pair is their SUM, $a+l$, not their equality.
The derivation ends with $2 S_n = n[2a+(n-1)d]$. Why divide both sides by $2$ at this last step?
because $n$ is always an even number
because $d$ must be halved before it can be used in the formula
because only half the terms were actually added together
forward-plus-backward counts every term twice
Check your answer
because $n$ is always an even number — The derivation works for any whole number $n$ — the division by $2$ comes from the doubled counting, not from any property of $n$ itself.
because $d$ must be halved before it can be used in the formula — $d$ is never halved in this formula — it is the SUM, counted twice over by adding forward and backward, that gets halved back down.
because only half the terms were actually added together — Every term appears in BOTH the forward and backward sums — none are skipped; the halving corrects for the doubling, not for missing terms.
✓ forward-plus-backward counts every term twice — (D) $2S_n$ is the sum counted twice over, once forward and once backward — dividing by $2$ removes that doubling to leave the true $S_n$.
A student derived $S_n = n[2a+(n-1)d]$, without the final division by 2. What is wrong?
this is exactly double the correct sum
nothing is wrong — this is a correct alternate form of the sum formula
the error is that $(n-1)$ should be $n$ instead
Check your answer
✓ this is exactly double the correct sum — (A) $n[2a+(n-1)d]$ is what the derivation calls $2S_n$ — it must still be divided by $2$ to reach the true $S_n = n/2 \cdot [2a+(n-1)d]$.
nothing is wrong — this is a correct alternate form of the sum formula — Without dividing by $2$, this expression is exactly twice the true sum — it is not an equivalent alternate form, but an uncorrected doubled value.
the error is that $(n-1)$ should be $n$ instead — The bracket, $2a+(n-1)d$, is already correct — the missing step is only the final division by $2$, not a change inside the bracket.
the monthly deposits form an AP with $a = 200$ and $d = 40$ The first month’s deposit is ₹200 and each later one is a fixed ₹40 more than the month before. It is the DEPOSITS that form the AP; the total is their sum.
$S_{18} = 9 \cdot [400 + 680] = 9 \cdot 1080$ Finish the arithmetic inside the bracket, then multiply.
$S_{18} = 9720$ Complete the multiplication.
the total saved after 18 months is ₹9720 *This asks for a running total, $S_n$, not the amount added in the 18th month alone, which would be $a_{18}$.*
$l = 200 + (18 - 1) \cdot 40 = 880$, then $S_{18} = 18/2 \cdot (200 + 880) = 9720$ Check the total a second way, using the $l$-form of the sum formula, where $l$ is the 18th month’s own deposit: it lands on ₹9720 again, the same total found above.
Each month’s deposit is ₹40 more than the last, and the six pair evenly above and below the dashed average line.
the first month
a little more every month
Kabir keeps each month’s coins in his savings box, and every month’s stack is a little taller than the one before.
Find the total saved, when a fixed amount grows by a fixed amount every month.
Find a and d The first month’s deposit is $a$. The fixed amount each later deposit adds on top of the one before is $d$.
Write the sum formula Use $S_n = n/2 \cdot [2 a + (n - 1) d]$ for the number of months given.
Do the arithmetic Work out the bracket first, then multiply.
Read it as a total, not a term The answer is everything saved so far, not the amount added in the last month alone.
Check yourself
A savings box takes a deposit of ₹150 in the first month, and each following month’s deposit is ₹30 more than the one before it. Using the same method as this chapter’s savings example, find the total saved after 12 months.
₹480
₹3780
₹5760
₹4140
Check your answer
₹480 — ₹480 is the single deposit figure for month 12 alone — the question asks for the TOTAL saved across all 12 months, which is $S_{12}$.
₹5760 — Multiplying one month’s figure by 12 assumes every month deposited the same amount — but the deposit grows by ₹30 each month, so the months are not all equal.
₹4140 — The bracket needs $(n-1) \cdot d$, which is $11 \cdot 30=330$, not $12 \cdot 30=360$ — using $n$ there overshoots by one step of $d$.
The same $S_n$ reasoning used for a savings box also answers ‘what is the total distance covered by a runner who covers 2 km in the first lap and 0.5 km more in each following lap, over 10 laps?’ Why does the same formula apply to both?
because both situations involve money, even though laps are measured in kilometres
both add a starting amount and a fixed extra amount at every repeated step
because both situations always run for exactly 12 steps
Check your answer
because both situations involve money, even though laps are measured in kilometres — Laps are measured in kilometres, not rupees — what actually connects the two situations is the fixed-step accumulation pattern, not a shared unit.
✓ both add a starting amount and a fixed extra amount at every repeated step — (B) A savings box and a set of laps both grow by a fixed amount each time, from a fixed starting value — the sum formula measures the total of any such repeating, fixed-step accumulation.
because both situations always run for exactly 12 steps — The two scenarios use different counts, 12 months and 10 laps — the shared feature is the fixed-step accumulation, not a matching number of steps.
For the chapter’s own savings example ($a=200$, $d=40$, 18 months), a student answered ‘the total saved’ as $a_{18} = 200+(18-1) \cdot 40=880$. What went wrong?
$880$ is month 18’s deposit alone, not the running total
nothing is wrong — $880$ correctly answers the total saved
the error is only in the arithmetic of $18-1$, not in which formula was used
Check your answer
✓ $880$ is month 18’s deposit alone, not the running total — (A) ‘Total saved’ asks for the running sum of every month’s deposit, $S_{18}=9720$, not the single deposit made in month 18 alone, $a_{18}=880$.
nothing is wrong — $880$ correctly answers the total saved — $880$ is only month 18’s own deposit — the total across all 18 months adds up every deposit, giving $9720$, a different and larger number.
the error is only in the arithmetic of $18-1$, not in which formula was used — The arithmetic inside $a_{18}$ is correct — the real error is reaching for the TERM formula when the question asked for a running total, which needs $S_n$.
$a = 16$, $d = -2$, $S_n = 60$ Read the first term, common difference, and target sum off the AP and the question.
$60 = n/2 \cdot [2 \cdot 16 + (n - 1) \cdot (-2)]$ Substitute the known values into $S_n = n/2 \cdot [2 a + (n - 1) d]$.
$n^2 - 17 n + 60 = 0$ Expand and simplify, a quadratic in $n$.
$(n - 5) (n - 12) = 0$ Factorise the quadratic.
$n = 5$ or $n = 12$ Either factor can be zero.
both answers are valid *Since $a$ is positive and $d$ is negative, the terms from the 6th through the 12th add to zero, so the sum reaches $60$ twice.*
$S_5 = 5/2 \cdot [32 + 4 \cdot (-2)] = 60$ and $S_{12} = 12/2 \cdot [32 + 11 \cdot (-2)] = 60$ Check both answers by substituting back into the sum formula: $n = 5$ and $n = 12$ both give a sum of $60$.
The last seven bars add to zero, so the first five bars and the first twelve bars reach the same total.
Find how many terms of an AP must be added to reach a given sum.
Read off a, d and the target sum Take these straight from the AP and the question.
Substitute into the sum formula Put $a$, $d$ and $S_n$ into $S_n = n/2 \cdot [2 a + (n - 1) d]$.
Simplify to a quadratic Expand and collect terms to get a quadratic in $n$.
Factorise and solve Find the two values of $n$ that make the quadratic zero.
Check both answers Substitute each value of $n$ back into the sum formula to see if both are valid.
Check yourself
Solving $S_n=n/2 \cdot [2a+(n-1)d]$ for $n$, given $a$, $d$ and $S_n$, generally produces what kind of equation in $n$?
a linear equation, since $n$ appears only once in the formula
an equation with no solution for $n$ at all
$(n-1)$ times $n$ in the formula gives an $n^2$ term
an equation only involving $d$, with $n$ cancelling out
Check your answer
a linear equation, since $n$ appears only once in the formula — $n$ appears twice, once outside the bracket and once inside it as $(n-1)$ — multiplying them together produces an $n^2$ term, not a linear one.
an equation with no solution for $n$ at all — The equation is solvable — it is a quadratic in $n$, which can be solved by factoring or the quadratic formula, just like any other.
✓ $(n-1)$ times $n$ in the formula gives an $n^2$ term — (C) Expanding $n/2 \cdot [2a+(n-1)d]$ multiplies $n$ by the bracket containing $(n-1)d$, which produces an $n^2$ term — a quadratic in $n$.
an equation only involving $d$, with $n$ cancelling out — $n$ is the very quantity being solved for — it does not cancel out; it remains in the equation as the unknown of a quadratic.
An AP begins $20, 17, 14, \cdots$. How many terms must be added to give a sum of $65$?
$6$
$4$
$5$
$65 / 20 = 3.25$, rounded to $3$
Check your answer
$6$ — Adding a 6th term overshoots the target — the sum of the first 5 terms alone already reaches $65$.
$4$ — The first 4 terms only add to $20+17+14+11=62$, short of $65$ — one more term is needed to reach it.
✓ $5$ — (C) $a=20$, $d=-3$; checking directly, $20+17+14+11+3=65$, so 5 terms are needed.
$65 / 20 = 3.25$, rounded to $3$ — Dividing the target sum by the first term is not how $n$ is found — $n$ comes from solving the actual sum formula, not from this shortcut.
Solving $S_n=60$ for the AP $16, 14, 12, \cdots$ correctly gives $n=5$ or $n=12$. A student then reported the number of terms as $6$, ‘one more than the smaller solution, to be safe’. What is wrong?
nothing is wrong — rounding up by one term is a safe way to report the answer
the error is that only $n=12$ can ever be correct, never $n=5$
$n$ must be exactly $5$ or $12$ — nothing else
Check your answer
nothing is wrong — rounding up by one term is a safe way to report the answer — Only $n=5$ and $n=12$ actually satisfy $S_n=60$ for this AP — adjusting to $6$ is not a safe rounding, it is simply an incorrect answer.
the error is that only $n=12$ can ever be correct, never $n=5$ — Both $n=5$ and $n=12$ are genuinely valid here, since the middle terms from 6 through 12 cancel to zero — the actual error is the invented ‘one more’ adjustment, not a need to discard either solution.
✓ $n$ must be exactly $5$ or $12$ — nothing else — (C) The equation’s solutions are $n=5$ and $n=12$ exactly — no other value of $n$ satisfies $S_n=60$, so adjusting the count ‘to be safe’ produces a wrong answer.
COMMON DIFFERENCE: $d$ is the fixed step from one term to the next. From $5$ to $11$, $d = 11 - 5 = 6$.
NTH TERM: $a_n = a + (n - 1) d$. The 4th term of that same AP is $a_4 = 5 + (4 - 1) \cdot 6 = 23$.
SUM: $S_n = n/2 \cdot [2 a + (n - 1) d]$. Its first four terms add to $S_4 = 4/2 \cdot [10 + 18] = 56$.
TERM VS SUM: $a_n = S_n - S_{n - 1}$. Here $a_4 = 56 - 33 = 23$, matching the term above.
Let us step back over what we have built. Every arithmetic progression is fixed by exactly two numbers: the first term $a$ and the common difference $d$. Once you know both, nothing else is needed to write out the whole list.
From those two numbers, two formulas follow, and they answer two different questions. The $n$th term, $a_n = a + (n - 1) d$, is one single value: a place on the list, and nothing more. The sum of the first $n$ terms is $S_n = n/2 \cdot [2 a + (n - 1) d]$. It is a running total: everything added together, up to that point.
A salary in a given year is one term. Total savings after several months is a sum. Before you answer a question about an AP, check which one it is asking for. Your turn: a question asks for the total after 4 years. Term, or sum? (Answer: sum, $S_4$.) Confusing the two sends a calculation to the wrong formula, no matter how carefully the arithmetic is done afterward.
Check yourself
Which correctly summarises how $a_n$ and $S_n$ relate, for the same AP?
$a_n$ and $S_n$ are two different formulas for the same quantity
$a_n$ is always larger than $S_n$ for the same AP
$S_n$ only exists for infinite APs, never finite ones
one term, one running total, linked only by subtraction
Check your answer
$a_n$ and $S_n$ are two different formulas for the same quantity — $a_n$ and $S_n$ measure different things entirely, a single term and a running total — they are never the same quantity, however computed.
$a_n$ is always larger than $S_n$ for the same AP — No fixed size relationship holds between $a_n$ and $S_n$ in general — for many APs, especially with negative early terms, $S_n$ can be larger, smaller, or even negative itself.
$S_n$ only exists for infinite APs, never finite ones — $S_n$ sums exactly $n$ terms, a finite count — it is defined for finite APs (and for a finite stretch of an infinite one), not restricted to infinite APs alone.
✓ one term, one running total, linked only by subtraction — (D) This captures all three facts at once: $a_n$ is one value, $S_n$ is a total, and the only bridge between them is the subtraction formula, never equality.
Thinking of an AP as evenly spaced points along a straight line helps explain which fact about $d$?
$d$ can be positive, negative, or zero, since a line can rise, fall, or stay flat
$d$ must always be a whole number, since points on a line sit at whole-number marks
$d$ must always be positive, since a line always moves forward
$d$ has no connection to the line picture at all
Check your answer
✓ $d$ can be positive, negative, or zero, since a line can rise, fall, or stay flat — (A) Points evenly spaced along a line can rise, fall, or stay level — matching exactly how $d$ can be positive, negative, or zero.
$d$ must always be a whole number, since points on a line sit at whole-number marks — A line has points at every value, not just whole numbers — $d$ can be any real number at all, not restricted to whole numbers.
$d$ must always be positive, since a line always moves forward — A line can descend or stay perfectly flat just as easily as it can rise — none of those directions is ruled out for $d$.
$d$ has no connection to the line picture at all — The line picture is exactly what explains why $d$ can take any sign — evenly spaced points can rise, fall, or stay flat, just as $d$ can be positive, negative, or zero.
A stadium has 30 rows. Row 1 seats 40 people, and each row behind it seats 3 more than the row in front. How many people does row 30 seat, and how many people does the whole stadium seat in total?
row 30 seats $127$; the stadium seats $127$ in total
row 30 seats $130$; the stadium seats $2505$ in total
row 30 seats $127$; the stadium seats $3810$ in total
row 30 seats $127$; the stadium seats $2505$ in total
Check your answer
row 30 seats $127$; the stadium seats $127$ in total — $127$ correctly answers row 30 alone — the stadium’s TOTAL across all 30 rows is a much larger running sum, $2505$, not a repeat of one row’s figure.
row 30 seats $130$; the stadium seats $2505$ in total — Row 30 needs only $29$ increments of $3$ from row 1, not $30$ — that extra increment overshoots the correct row-30 figure of $127$.
row 30 seats $127$; the stadium seats $3810$ in total — Multiplying row 30’s seating by 30 assumes every row seats the same number of people — but seating grows by 3 each row, so the rows are not all equal.
✓ row 30 seats $127$; the stadium seats $2505$ in total — (D) $a_{30}=40+(30-1) \cdot 3=40+87=127$; $S_{30}=30/2 \cdot [2 \cdot 40+(30-1) \cdot 3]=15 \cdot [80+87]=15 \cdot 167=2505$.
Check yourself: the whole chapter
The first term of a sequence is 7, and 5 is added to each term to get the next one. Which of these is the correct start of this sequence?
7, 35, 175, 875
7, 12, 17, 22
7, 2, -3, -8
7, 12, 24, 48
Check your answer
7, 35, 175, 875 — Adding 5 each time means 5 is a step to be added, not a factor to multiply by — multiplying builds a geometric progression, not this arithmetic one.
✓ 7, 12, 17, 22 — (B) Adding the fixed amount 5 to each term in turn gives 7, 12, 17, 22 — the step never changes.
7, 2, -3, -8 — The common difference here is added, not subtracted — 7 + 5 = 12, not 7 - 5 = 2.
7, 12, 24, 48 — The step from term to term must stay the same fixed number throughout — here the first step is 5, but the later steps grow to 12 and then 24, so the sequence is not built with a constant step.
An AP has first term $a = 4$ and common difference $d = 6$. Using $a_n = a + (n - 1) d$, what is the 5th term?
34
10
28
22
Check your answer
34 — The formula needs (n - 1) additions of d, not n — the first term itself needs zero additions, so reaching the 5th term needs only 4 steps, not 5.
10 — Every term after the first needs its own addition of d — reaching the 5th term needs d added 4 times, 6 dot 4 = 24, not just once.
Which of these lists of numbers is an arithmetic progression?
2, 4, 8, 16, 32
3, 6, 9, 13
5, 8, 11, 14
10, 7, 7, 4
Check your answer
2, 4, 8, 16, 32 — 4 - 2 = 2 but 8 - 4 = 4 — the differences are not equal, even though each term is double the one before it; a constant ratio is a different pattern (a geometric progression), not an AP.
3, 6, 9, 13 — 6 - 3 = 3 and 9 - 6 = 3 look promising, but 13 - 9 = 4 breaks the pattern — a list must be checked across every consecutive pair, not just the first one, before it can be called an AP.
✓ 5, 8, 11, 14 — (C) Every consecutive difference here is 3 — 8 - 5 = 3, 11 - 8 = 3, 14 - 11 = 3 — so this is an AP with common difference 3.
10, 7, 7, 4 — 7 - 10 = -3, then 7 - 7 = 0 — the step changes partway through, so this list has no single common difference.
In an AP, the 3rd term is 13 and the 7th term is 29. Using $a_n = a + (n - 1) d$, what is the first term $a$?
13
5
1
21
Check your answer
13 — The 3rd term is $a_3$, not $a_1$ — the first term still has to be found using $a_n = a + (n - 1) d$, it is not simply whichever term happens to be given.
✓ 5 — (B) $a + 2 d = 13$ and $a + 6 d = 29$; subtracting gives $4 d = 16$, so $d = 4$, and $a = 13 - 2 \cdot 4 = 5$.
1 — The correct equations use (n - 1), so the 3rd term gives $a + 2 d = 13$, not $a + 3 d = 13$ — using the term number itself shifts every answer for $a$.
21 — Subtracting the first equation from the second, 29 - 13, gives $4 d = 16$ and $d = 4$; subtracting in the other order flips the sign and gives a wrong value for both $d$ and $a$.
You will meet a fixed step almost every month of your life: a price, a saving, a distance. Here are seven of those places.
Tracking a rising allowance. Your pocket money is $₹200$ this month. It goes up by a fixed $₹50$ every month after that. The maths: the fixed $₹50$ added each time is the common difference of the AP. From month $1$ to month $2$: $250 - 200 = 50$. From month $2$ to month $3$: $300 - 250 = 50$.
Checking if stair heights are even. A staircase has $6$ steps at heights $15, 30, 45, 60, 75, 90$ cm from the ground. The maths: the list is an AP only if the gap between consecutive heights stays the same throughout. Gaps: $30 - 15 = 15$, $45 - 30 = 15$, $60 - 45 = 15$. Every gap is $15$ cm.
Pricing a tailor’s extra pieces. A tailor’s bill is $₹150$ for one piece, and $₹20$ more for each extra piece. Two pieces cost $₹170$ and three cost $₹190$. The maths: the bill for $n$ pieces is the $n$th term of an AP, $a_n = a + (n - 1) d$. For $10$ pieces: $a_{10} = 150 + (10 - 1) \cdot 20 = 330$ rupees.
Adding up ten days of jogging. You jog $2$ km on day $1$. You add $0.5$ km more each day after that. The maths: the total distance over $n$ days is the sum of the first $n$ terms of an AP. Over $10$ days: $S_{10} = 10/2 \cdot [2 \cdot 2 + 9 \cdot 0.5] = 5 \cdot 8.5 = 42.5$ km.
Reading a savings app’s running total. Your savings app shows a running total of $₹1350$ after $6$ deposits and $₹1550$ after $7$. The maths: a single deposit is a term, found as the difference between two consecutive running sums. The $7$th deposit: $a_7 = 1550 - 1350 = 200$ rupees.
Any single term of an AP is the difference between two neighbouring sums, so a running total is enough to recover it.
Finding a print shop’s rate from two invoices. A print shop’s price list is an AP. Printing $4$ posters costs $₹700$, and printing $9$ costs $₹1200$. The maths: two known terms give two equations in $a$ and $d$, and solving them finds both. $a + 3d = 700$ and $a + 8d = 1200$. Subtracting: $5d = 500$, so $d = 100$ and $a = 400$.
Counting bib numbers from the back. A marathon assigns bib numbers as an AP: $101, 103, 105, \dots$ up to $199$ for the last runner. The maths: counting from the end reverses the AP. The $r$th bib from the back uses the last term as the new start. $5$th bib from the end: $l + (r - 1)(-d) = 199 + 4 \cdot (-2) = 191$.
Your turn. A sapling is $20$ cm tall now. It grows a fixed $4$ cm every week. How tall will it be after $12$ more weeks? Answer: Term $13$ (now plus $12$ weeks): $a_{13} = 20 + (13 - 1) \cdot 4 = 68$ cm.
practice Is $7, 11, 15, 19, \cdots$ an AP? If so, find $d$ and the next two terms. (Worked in full below — read it, then do the next two the same way.)
practice Is $2, 6, 18, 54, \cdots$ an AP? Give a reason. (Start the same way — subtract each term from the one after it, and compare the gaps.)
practice Write the first four terms of the AP with $a = 6$ and $d = -4$. (The first term is $a$; each term after it adds $d$.)
practice A taxi charges ₹25 for the first kilometre and ₹8 for each additional kilometre after that. Do the total fares for 1, 2, 3, 4 kilometres form an AP? If so, what is $d$?
practice Write the first four terms of the AP with $a = 10$ and $d = -3$.
practice For the AP $2, 5, 8, 11, \cdots$, write $a$ and $d$.
practice Is $3, 6, 12, 24, \cdots$ an AP? Give a reason.
practice Is $-1, -3, -5, -7, \cdots$ an AP? If so, find $d$ and the next two terms.
practice A hall has 30 seats in the first row, 32 in the second, 34 in the third, and so on. Do the number of seats per row form an AP? If so, find $d$.
Answers
Yes, it is an AP with $d = 4$. The next two terms are $23$ and $27$.
No. The gaps are $4$, $12$ and $36$, which are not equal, so the list is not an AP.
$6, 2, -2, -6$.
Yes — the fares are ₹25, ₹33, ₹41, ₹49, an AP with $d = 8$.
$10, 7, 4, 1$.
$a = 2$, $d = 3$.
No — the differences $6 - 3 = 3$ and $12 - 6 = 6$ are not equal.
Yes, $d = -2$; the next two terms are $-9$ and $-11$.
Yes, an AP with $d = 2$.
Exercise 5.1 — further practice
practice Write the first four terms of the AP with $a = -7$ and $d = 5$.
practice The cost of digging a well is ₹150 for the first metre and rises by ₹50 for each additional metre after that. Do the costs of digging 1, 2, 3, 4 metres form an AP? If so, find $d$.
practice A tank holds 6000 litres of water. Every hour, 20% of the water currently in the tank is drained out. Do the amounts of water left in the tank at the end of each hour form an AP? Give a reason.
practice Vikram saves ₹500 in the first month of a year and increases his saving by a fixed ₹200 every month after that. Do his monthly savings for the first four months form an AP? If so, find $d$.
practice Is $46, 39, 32, 25, \cdots$ an AP? If so, find $d$ and the next two terms.
practice Is $5, 10, 20, 40, \cdots$ an AP? Give a reason.
practice For the AP $-0.8, -0.3, 0.2, 0.7, \cdots$, write $a$ and $d$, and find the next term.
Answers
$-7, -2, 3, 8$.
Yes — the costs are ₹150, ₹200, ₹250, ₹300, an AP with $d = 50$.
No — the amount drained each hour depends on how much water remains (a fixed fraction, not a fixed amount), so the differences between successive amounts are not equal.
Yes — the savings are ₹500, ₹700, ₹900, ₹1100, an AP with $d = 200$.
Yes, $d = -7$; the next two terms are $18$ and $11$.
No — $10 - 5 = 5$ but $20 - 10 = 10$; the differences are not equal, since each term is double the one before it, not a fixed amount more.
practice In an AP, $a = 4$, $d = 6$ and $n = 10$. Find $a_n$. (Worked in full below — read it, then do the next two the same way.)
practice In an AP, $a = 3$, $d = 7$ and $a_n = 59$. Find $n$. (Start the same way — write $a_n = a + (n - 1) d$ and put in what you know. This time $n$ is what is missing.)
practice Which term of the AP $5, 9, 13, \cdots$ is $101$? (Read $a$ and $d$ off the list first.)
practice In an AP, $a = 7$, $d = 3$, $n = 8$. Find $a_n$.
practice In an AP, $a = 5$, $d = 4$, and $a_n = 45$. Find $n$.
practice Which term of the AP $21, 18, 15, \cdots$ is $-81$?
practice Find three numbers between $8$ and $28$ that, together with $8$ and $28$, form an AP of five terms.
practice The 3rd term of an AP is $12$ and the 8th term is $37$. Find $a$ and $d$.
practice A finite AP has 60 terms. Its last term is $10$ and its common difference is $-2$. Find the 5th term from the end.
practice A company’s monthly water bill starts at ₹1200 and rises by a fixed ₹150 every month. Find the bill in the 10th month.
Answers
$a_{10} = 58$.
$n = 9$.
The 25th term.
$a_8 = 7 + (8 - 1) \cdot 3 = 28$.
$45 = 5 + (n - 1) \cdot 4$, so $n - 1 = 10$ and $n = 11$.
practice Find the sum of the first 12 terms of the AP $4, 9, 14, \cdots$ (Worked in full below — read it, then do the next two the same way.)
practice In an AP, $a = 6$, $d = 3$ and $S_n = 195$. Find $n$. (Start the same way — write the sum formula and substitute. What is left is a quadratic in $n$.)
practice A person saves ₹20 in the first week and ₹5 more in each following week. Find the total saved over 10 weeks. (Name $a$, $d$ and $n$ from the words before reaching for the formula.)
practice Find the sum of the first 15 terms of the AP $3, 7, 11, \cdots$
practice In an AP, $a = 8$, $d = 5$, and $S_n = 305$. Find $n$.
practice How many terms of the AP $24, 21, 18, \cdots$ must be taken to give a sum of $78$?
practice A person saves ₹32 in the first week and ₹4 more every following week. Find the total savings over 12 weeks.
practice A construction contract has a penalty clause: ₹200 for the first day of delay, increasing by ₹50 for each additional day. Find the total penalty for a 10-day delay.
practice A prize fund of ₹7000 is to be split among 7 winners so that each winner receives ₹100 less than the winner ranked just above. Find the amount the first-ranked winner receives.
practice Section A of a class plants 3 saplings on the first day of a plantation drive and 2 more each following day. Find the total number of saplings Section A plants in 5 days.
$a = 24$, $d = -3$. $78 = n/2 \cdot [48 - 3 (n - 1)]$ simplifies to $n^2 - 17 n + 52 = 0$, which factors as $(n - 4) (n - 13) = 0$; both $n = 4$ and $n = 13$ are valid, since $a$ is positive and $d$ is negative and the terms between them cancel.
practice Find the sum of the first 20 terms of the AP $5, 9, 13, \cdots$
practice In an AP, $a = 4$, $d = 5$, and $S_n = 216$. Find $n$.
practice In an AP, $d = 4$, $n = 12$, and $S_n = 288$. Find $a$.
practice In an AP, $S_{10} = 250$ and $S_9 = 207$. Find $a_{10}$.
practice The sum of the first 10 terms of the AP $2, 4, 6, 8, \cdots$ is
$90$
$100$
$110$
$120$
practice The first term of an AP is $-5$, the last term is $46$, and the sum is $205$. Find the number of terms.
practice Find the sum of the first 25 terms of the AP $40, 35, 30, \cdots$
practice How many terms of the AP $20, 18, 16, \cdots$ must be taken to give a sum of $90$?
practice A charity fair sells tickets over several days: 20 tickets on the first day, and 5 more each day than the day before. If 570 tickets are sold in total, how many days did the sale run?
practice Bricks are stacked in rows so that there are 24 bricks in the bottom row, and each row above has 2 fewer bricks than the row below it. If the stack has 9 rows, find the number of bricks in the top row and the total number of bricks in the stack.
practice A swimming pool is drained so that 500 litres are removed in the first minute, and each following minute 40 fewer litres are removed than the minute before, until a total of 2400 litres have been removed. Find the number of minutes needed, given that the amount removed each minute cannot be negative.
Answers
$S_{20} = 860$.
$n = 9$.
$a = 2$.
$a_{10} = 43$.
C — $110$.
$n = 10$.
$S_{25} = -500$.
$n = 6$ or $n = 15$ — both are valid, since $a$ is positive and $d$ is negative and the terms between them cancel.
$12$ days.
The top row has $8$ bricks; the stack has $144$ bricks in total.
$n = 6$ minutes — the other root, $n = 20$, is rejected because it would require removing a negative amount of water in some minute.
practice An AP has $a = 60$ and $d = -7$. Find its first negative term. (Worked in full below — read it, then do the next two the same way.)
practice An AP has $a = 90$ and $d = -8$. Find its first term that is less than $20$. (Start the same way — simplify $a_n$ first, then ask which $n$ makes it drop below $20$.)
practice In an AP, $a_4 = 17$ and $a_9 = 42$. Find $a$ and $d$. (Write each of the two terms with the formula, then subtract one equation from the other.)
practice An AP has $a = 105$ and $d = -6$. Find its first negative term.
practice A ladder has 11 rungs. The bottom rung is 2500 mm wide and the top rung is 1000 mm wide, with the width decreasing uniformly rung by rung. Find the total length of wood needed for all the rungs together.
practice House numbers on one side of a street run $1, 2, 3, \cdots, 49$. Find the house number $x$ such that the sum of the house numbers before $x$ equals the sum of the house numbers after it.
practice A stepped terrace has 15 steps. The first step is 25 cm above the ground, and each step is 25 cm higher than the one before it. Find the height of the top step above the ground.
practice In an AP, the sum of the first 7 terms is $49$ and the sum of the first 17 terms is $289$. Find the sum of the first $n$ terms.
Answers
The 10th term, $a_{10} = -3$.
The 10th term, $a_{10} = 18$.
$d = 5$ and $a = 2$.
$a_n = 105 - 6 (n - 1)$; setting $a_n < 0$ gives $n > 18.5$, so $n = 19$ and $a_{19} = 105 - 6 \cdot 18 = -3$ is the first negative term.
$S_{11} = 11/2 \cdot (2500 + 1000) = 19250$, in millimetres — 19.25 metres of wood.
The total of all 49 house numbers is $1225$; setting the sum before $x$ equal to the sum after $x$ gives $x^2 = 1225$, so $x = 35$.
$a_{15} = 25 + (15 - 1) \cdot 25 = 375$; the top step is 375 cm (3.75 m) above the ground.
Solving $a + 3 d = 7$ and $a + 8 d = 17$ gives $d = 2$ and $a = 1$; then $S_n = n/2 \cdot [2 + 2 (n - 1)] = n^2$.
Exercise 5.4 (Optional) — further practice
practice In an AP, $a_3 = 10$ and $a_7 = 26$. Which of these is its common difference $d$?
$3$
$4$
$5$
$6$
practice An AP has $a = 88$ and $d = -7$. Find its first term that is less than $10$.
practice A hall has 25 seats in the front row, and each row behind has 3 more seats than the row in front. There are 18 rows in the hall. Find the number of seats in the last row and the total seating capacity of the hall.
practice A supermarket stacks tin cans in a pyramid display with 15 cans in the bottom row, 2 fewer cans in each row above it, and a single can on top. How many rows are there, and how many cans are used in total?
practice The sum of the 4th and 8th terms of an AP is $20$, and their product is $96$. Find the first three terms of the AP, given that the AP is increasing.
practice In an AP, the sum of the first 6 terms is $72$ and the sum of the first 16 terms is $512$. Find the sum of the first $n$ terms.
Answers
B — $4$.
$a_{13} = 4$.
The last row has $76$ seats; the hall seats $909$ in total.
$8$ rows; $64$ cans in total.
$5, 6, 7$.
$S_n = 2 n^2$.
The five houses before house six add to fifteen, and the two houses after it add to fifteen too.
Seven marked points, one for each term count from 1 to 7, all sit exactly on one smooth upward curve.
