Arithmetic Progressions

Reena gets a job at a starting monthly salary of 8000 rupees, with a fixed annual increment of 500 rupees. A ladder's rungs shrink uniformly by 2 cm as you climb it. A school gives cash prizes of 200, 250, 300, and so on up to the topper of Class XII. Three completely different situations, and yet each one produces the same kind of list of numbers: a list where every term is obtained from the one before it by adding a single fixed number, over and over. This chapter names that pattern an arithmetic progression, or AP, and builds exactly two tools for handling it without ever writing out a hundred terms by hand. The first tool answers "what is the value at position n": the nth term formula, $a_n = a + (n-1)d$ , which turns Reena's 25th-year salary or the 30th two-digit multiple of 3 into a single substitution. The second tool answers "what do all the terms add up to": the sum formula, $S_n = \tfrac{n}{2}[2a+(n-1)d]$ , discovered in spirit by a ten-year-old Gauss reversing and re-adding a list of numbers, and just as quick whether $n$ is 10 or 10 million. Together these two formulas convert tedious, repetitive counting into two or three lines of algebra. This article follows the NCERT chapter's own path: first pin down what an AP actually is and how to test whether a given list qualifies, then derive and use the nth-term formula on salaries, temperatures, and "which term is this," then derive and use the sum formula on prize money, penalties, and stacked logs. Along the way you will meet the recurring traps that cost the most marks — a sign flipped in computing $d$ , an $n$ swapped for $n-1$ , a valid second root thrown away — and see why they happen and how to catch them every time. By the end, a jumble like $17, 21, 25, 29, \ldots$ up to $n = 40$ terms should look completely tame: name $a$ and $d$ in five seconds, and every other question about it becomes one formula away.

Key takeaways

An arithmetic progression (AP) is a list of numbers in which every term, except the first, is obtained by adding a fixed number to the term before it; that fixed number is the common difference $d$ , and it can be positive, negative, or zero.
The general form of an AP is $a,\ a+d,\ a+2d,\ a+3d,\ \ldots$ , where $a$ is the first term.
Testing for an AP: a list $a_1, a_2, a_3, \ldots$ is an AP exactly when $a_{k+1}-a_k$ is the same value for every $k$ ; checking only one pair is not enough.
Always compute $d$ as (later term) minus (earlier term), $d = a_{k+1}-a_k$ , never the reverse — the sign of $d$ carries real meaning.
The nth term (general term) is $a_n = a + (n-1)d$ ; it lets you find any term, or check whether a given number is a term at all by solving for $n$ and requiring it to be a positive integer.
A finite AP has a last term $l = a_m$ ; the term counted $k$ places from the end equals $l - (k-1)d$ , the mirror image of counting from the start.
The sum of the first n terms is $S_n = \dfrac{n}{2}\left[2a+(n-1)d\right]$ , which can also be written $S_n = \dfrac{n}{2}(a+a_n)$ or, using the last term, $S_n = \dfrac{n}{2}(a+l)$ .
Solving $S_n =$ (given value) for $n$ produces a quadratic in $n$ and can legitimately give two valid positive-integer roots at once.
If $a, b, c$ are three consecutive terms of an AP, $b$ is their arithmetic mean, $b = \dfrac{a+c}{2}$ .
Key idea: an arithmetic progression is a list built by repeatedly adding a fixed common difference d to a first term a, so any term is found instantly by an = a + (n-1)d and the sum of the first n terms by Sn = n/2[2a + (n-1)d], turning problems that look like tedious counting into two or three lines of algebra.

The core insight

An arithmetic progression is a list of numbers $a, a+d, a+2d, \ldots$ built by repeatedly adding a fixed common difference $d$ to a first term $a$ , so that the general term is $a_n = a+(n-1)d$ and the sum of the first $n$ terms is $S_n = \dfrac{n}{2}[2a+(n-1)d] = \dfrac{n}{2}(a+l)$ , letting any term or any sum be found by direct substitution rather than by writing out and adding every term one at a time.

What this chapter covers

The chapter builds two formulas on top of one careful definition, in a fixed order that mirrors the sections below.

Section 5.2 defines an arithmetic progression from everyday examples — a salary with a fixed increment, a ladder's shrinking rungs, a queue of students' heights — and gives the precise test, $a_{k+1}-a_k$ constant, for deciding whether a list of numbers is an AP.
Section 5.3 derives the nth term formula $a_n = a+(n-1)d$ by unrolling the "add $d$ each time" rule step by step, then applies it to finding a term, checking membership, finding an unknown AP from two given terms, and counting terms from the end.
Section 5.4 derives the sum of the first n terms, $S_n = \tfrac{n}{2}[2a+(n-1)d]$ , using the reverse-and-add trick attributed to the young Gauss, then applies it to savings, production, and "how many terms give this sum" problems that turn into quadratics.
A closing note connects three consecutive AP terms to the arithmetic mean, $b = (a+c)/2$ .

Worked examples below reuse NCERT's own example and exercise numbers wherever named directly, so you can cross-check against your textbook.

The core ideas

Defining an arithmetic progression

Look at the list $6, 9, 12, 15, \ldots$ . Each term after the first is obtained by adding the fixed number $3$ to the one before it: $9 = 6+3$ , $12 = 9+3$ , $15 = 12+3$ . A list built this way is called an arithmetic progression (AP), and the fixed number added at every step is its common difference, written $d$ .

a,\ a+d,\ a+2d,\ a+3d,\ \ldots

where: $a$ is the first term of the AP and $d$ is the common difference, found from any two consecutive terms as $d = a_{k+1}-a_k$ . This is the general form of an AP — every AP can be written this way once $a$ and $d$ are known. Crucially, $d$ need not be positive: for $6, 3, 0, -3, \ldots$ , $d = 3-6 = -3$ ; and $d$ can even be zero, as in $2, 2, 2, 2, \ldots$ , technically an AP with $d=0$ . Knowing just $a$ or just $d$ alone is never enough — an AP is fixed only once both are known.

Testing whether a list of numbers forms an AP

Given only a list of numbers, how do you decide whether it is an AP at all? The rule is to compute the difference between every pair of consecutive terms and check that this difference is the same value throughout the list, not merely for one pair.

d = a_{k+1} - a_k \quad \text{for every } k = 1, 2, 3, \ldots

where: $a_{k+1}$ and $a_k$ are the $(k+1)$ th and $k$ th terms. Subtract the earlier term from the later one, in that order, every time — even when the later term is smaller, as in $6, 3, 0, -3, \ldots$ , where $d = 3-6=-3$ , not $6-3=3$ . For $1, 1, 2, 3, 5, \ldots$ , the differences are $0, 1, 1, 2, \ldots$ — not constant — so this list is not an AP, even though it looks orderly. One matching pair proves nothing; check at least three consecutive differences before declaring a list an AP.

Figure

A number line showing the arithmetic progression six nine twelve fifteen with equal jumps of three units marked between every pair of consecutive terms

The nth term, or general term, of an AP

Reena's monthly salary in successive years is $8000, 8500, 9000, 9500, \ldots$ , an AP with $a=8000$ and $d=500$ . Writing the salary for year 3 as $8000 + 2\times500$ , year 4 as $8000+3\times500$ , and year 5 as $8000+4\times500$ reveals the pattern: the salary for year $n$ needs exactly $(n-1)$ additions of $d$ to the first term, never $n$ additions, because the first term itself already accounts for year 1 with zero additions.

a_n = a + (n-1)d

where: $a_n$ is the nth term (also the general term) of the AP, $a$ is the first term, $d$ the common difference, and $n$ the position, a positive integer. If the AP is finite with $m$ terms, $a_m$ is also called the last term, written $l$ . This single formula answers three questions at once: given $n$ , find $a_n$ by substitution; given a value, solve for $n$ to check membership (accepting only a positive-integer $n$ ); and given two terms, solve two simultaneous equations in $a$ and $d$ to pin down the whole AP.

The sum of the first n terms

Reversing and adding a list against itself, exactly as ten-year-old Gauss is said to have summed $1$ to $100$ , gives a shortcut for the sum of any AP without adding term by term — full proof in the derivation below.

S_n = \frac{n}{2}\left[2a+(n-1)d\right] = \frac{n}{2}(a+a_n)

where: $S_n$ denotes the sum of the first $n$ terms (also written $S$ when $n$ is clear), and the second form replaces $2a+(n-1)d$ with $a+a_n$ . When the last term $l$ is known directly, the formula is often more convenient as $S_n = \tfrac{n}{2}(a+l)$ — useful when $d$ is awkward to compute but $a$ and $l$ are both given. A useful remark follows at once: since $S_n$ is the sum up to term $n$ and $S_{n-1}$ up to term $n-1$ , their difference is the missing term, $a_n = S_n - S_{n-1}$ .

Figure

A graph of Reena salary AP eight thousand eighty five hundred nine thousand ninety five hundred ten thousand plotted against the year number with every point lying exactly on one straight line

The arithmetic mean of three consecutive terms

If $a, b, c$ are three consecutive terms of an AP, the common difference gives $b - a = c - b$ , since both equal $d$ . Rearranging, $2b = a+c$ .

b = \frac{a+c}{2}

where: $b$ is called the arithmetic mean of $a$ and $c$ — the middle term of any three consecutive AP terms is always the plain average of its neighbours. This is a compact, exam-friendly fact: given the first and third of three consecutive AP terms, the middle one is found without ever computing $d$ separately.

Key derivations

Derivation AThe nth term formula of an arithmetic progression

The goal of this derivation is to obtain a direct formula for the nth term of an AP with first term $a$ and common difference $d$ , so that any term can be found by one substitution instead of by repeatedly adding $d$ from the first term all the way up to the $n$ th. The strategy is to write out the first few terms using the definition "each term equals the previous term plus $d$ ," notice how many copies of $d$ have accumulated at each step, and generalise that count to an arbitrary position $n$ .

Step 1. Start from $a_1=a$ and $a_{k+1}=a_k+d$ . Applying this once gives the second term $a_2 = a_1+d = a+d = a+(2-1)d$ .

Step 2. Apply the rule again for the third term: $a_3 = a_2+d = (a+d)+d = a+2d = a+(3-1)d$ — two additions of $d$ , one fewer than the position number $3$ .

Step 3. Repeat once more: $a_4 = a_3+d = (a+2d)+d = a+3d = a+(4-1)d$ — three additions, again one less than the position.

Step 4. Read off the pattern: reaching position $n$ from $a_1=a$ always takes exactly $(n-1)$ additions of $d$ , since the first term itself uses zero. This gives the general term $a_n = a+(n-1)d$ , valid for every positive integer $n$ .

Step 5. Verify against NCERT's own Example 3, the AP $2, 7, 12, \ldots$ with $a=2$ , $d=5$ . The formula predicts $a_{10} = 2+9(5) = 47$ , matching direct counting: $2,7,12,17,22,27,32,37,42,47$ . Verified: $a_{10}=2+9(5)=47$ for the AP $2,7,12,\ldots$ , matching term-by-term counting.

Go deeper · Derivation B · 1 minThe sum of the first n terms of an arithmetic progression

Derivation BThe sum of the first n terms of an arithmetic progression

The goal of this derivation is to find a closed formula for $S_n$ , the sum of the first $n$ terms of an AP, without ever adding the terms one by one, using the same reverse-and-add idea told of the young Gauss summing the integers from $1$ to $100$ . The strategy is to write the sum once forward and once in reverse, add the two copies term by term, and show every matched pair produces the same constant total.

Step 1. Write $S_n$ in natural forward order: $S_n = a + (a+d) + (a+2d) + \cdots + [a+(n-1)d]$ .

Step 2. Write the same sum in reverse order, last term to first: $S_n = [a+(n-1)d] + [a+(n-2)d] + \cdots + (a+d) + a$ — legitimate because reversing a sum never changes its value.

Step 3. Add the two lines term by term. Every matched pair sums to exactly $2a+(n-1)d$ : the first pair is $a + [a+(n-1)d]$ , the second is $(a+d)+[a+(n-2)d]$ , and so on — as one term in a pair loses a multiple of $d$ , its partner gains the same multiple back.

Step 4. There are $n$ such pairs, so $2S_n = n[2a+(n-1)d]$ . Dividing by $2$ gives $S_n = \dfrac{n}{2}[2a+(n-1)d]$ , valid for any AP and any positive integer $n$ .

Step 5. Verify against NCERT's own money-box problem: $a=100$ , $d=50$ , $n=21$ . The formula gives $S_{21} = \tfrac{21}{2}[200+1000] = \tfrac{21}{2}(1200) = 12600$ , matching the ₹12600 collected by the 21st birthday. Verified: $S_{21} = \tfrac{21}{2}(200+1000) = 12600$ for $a=100, d=50, n=21$ .

Go deeper · Derivation C · 1 minThe nth term counted from the end of a finite AP

Derivation CThe nth term counted from the end of a finite AP

The goal of this derivation is to find a term of a finite AP counting backward from its last term, without first working out the total number of terms and converting that into a forward position — a shortcut for the "kth term from the end" question the chapter poses for the AP $10, 7, 4, \ldots, -62$ . The strategy is to observe that a finite AP read backward is itself an AP, with the old last term $l$ as its new first term and $-d$ as its new common difference, then apply the ordinary nth-term formula to this reversed list.

Step 1. Take a finite AP $a_1, \ldots, a_m$ with first term $a$ , common difference $d$ , last term $l=a_m$ . Reversed, it reads $l, l-d, l-2d, \ldots$ , since going backward subtracts $d$ instead of adding it.

Step 2. This reversed list is an AP with first term $l$ and common difference $-d$ , so its $k$ th term is $l+(k-1)(-d) = l-(k-1)d$ .

Step 3. The $k$ th term of the reversed list is exactly the $k$ th term from the end of the original list. So the $k$ th term from the end equals $l-(k-1)d$ , with no need to first find the total number of terms.

Step 4. Verify against NCERT's own Example 8, the AP $10, 7, 4, \ldots, -62$ ( $a=10$ , $d=-3$ , $l=-62$ ), asked for the 11th term from the end: $l-(10)(-3) = -62+30 = -32$ .

Step 5. Cross-check the slow way: solving $-62=10+(n-1)(-3)$ gives $n=25$ total terms, so the 11th from the end is the $15$ th from the start ( $25-11+1$ ), and $a_{15}=10+14(-3)=-32$ — matching, and confirming "11th from the end" is the 15th term, not the 14th. Verified: both the reversed-AP shortcut and the direct count give $-32$ as the 11th term from the end of $10,7,4,\ldots,-62$ .

Worked examples

Board level - Which term equals a given value, and is any term zero

For the AP $21, 18, 15, \ldots$ , find which term equals $-81$ . Also determine whether any term of this AP is $0$ , giving a reason (this reuses NCERT's own Example 4).

Step 1. Identify $a=21$ and $d=18-21=-3$ . Use $a_n=a+(n-1)d$ with $a_n=-81$ , $n$ unknown.

Step 2. Substitute and solve: $-81 = 21+(n-1)(-3) = 24-3n$ , so $-105=-3n$ , giving $n=35$ — a positive integer, so $-81$ is genuinely a term.

Step 3. For "is any term $0$ ," set $a_n=0$ : $21+(n-1)(-3)=0$ gives $3(n-1)=21$ , so $n=8$ ; substituting confirms $a_8=21+7(-3)=0$ .

Step 4. Both answers come from the same formula, solved for different targets. The 35th term of the AP is $-81$ , and the 8th term is $0$ .

Board level - Total penalty as a sum of an AP

A construction contract specifies a penalty of ₹200 for the first day of delay, ₹250 for the second day, ₹300 for the third day, and so on, each succeeding day's penalty being ₹50 more than the day before. If the contractor delays the work by 30 days, how much money must be paid as penalty in total (this reuses NCERT's own Exercise 5.3, Question 15)?

Step 1. The penalties $200, 250, 300, \ldots$ form an AP with $a=200$ , $d=50$ , and $n=30$ terms, one per day of delay.

Step 2. The total penalty is the sum of these 30 terms: $S_{30}=\dfrac{30}{2}\left[2(200)+(29)(50)\right]$ .

Step 3. Simplify inside the brackets: $400+1450=1850$ , so $S_{30}=15\times1850$ .

Step 4. Multiply out: $15\times1850 = 27{,}750$ . The contractor must pay a total penalty of ₹27,750 for a 30-day delay.

Board HOTS / foundation JEE-NEET - An AP from a sum and a product of two terms

The sum of the third and seventh terms of an AP is $6$ , and their product is $8$ . Find the sum of the first sixteen terms of the AP (this reuses NCERT's own Exercise 5.4, Question 2).

Step 1. Translate into $a$ and $d$ : $a_3=a+2d$ , $a_7=a+6d$ . The sum condition gives $2a+8d=6$ , so $a+4d=3$ , giving $a=3-4d$ .

Step 2. Substitute into the product condition $(a+2d)(a+6d)=8$ : $(3-2d)(3+2d)=8$ .

Step 3. Expand as a difference of squares: $9-4d^2=8$ , so $d^2=\tfrac14$ , giving $d=\tfrac12$ or $d=-\tfrac12$ — two valid values, both to be carried through, not one discarded arbitrarily.

Step 4. For $d=\tfrac12$ : $a=1$ , so $S_{16}=8\left[2+7.5\right]=8(9.5)=76$ . For $d=-\tfrac12$ : $a=5$ , so $S_{16}=8\left[10-7.5\right]=8(2.5)=20$ . Both satisfy the conditions: $a=1,d=\tfrac12 \Rightarrow a_3=2,a_7=4$ ; $a=5,d=-\tfrac12 \Rightarrow a_3=4,a_7=2$ — sum $6$ , product $8$ either way. The sum of the first sixteen terms is $\mathbf{76}$ or $\mathbf{20}$ , depending on which valid AP is meant. Timing note: about four to five minutes. The trap is stopping after finding one value of $d$ — a quadratic in $d^2$ genuinely has two square roots, and both correspond to a real AP satisfying both given conditions, so both must be carried through to a final $S_{16}$ unless the question rules one out (for example, by specifying an increasing AP). Re-substitute both candidate APs into the original conditions before reporting either answer.

Common traps

Subtracting in the wrong order when finding d. Always compute $d = a_{k+1}-a_k$ ; for a decreasing AP like $6,3,0,-3,\ldots$ , subtracting backward silently flips the sign of $d$ .
Writing $a+nd$ instead of $a+(n-1)d$ . The first term uses zero additions of $d$ , so the count is always $(n-1)$ ; this slip appears in the nth-term, sum, and term-from-the-end formulas alike.
Checking only one pair of consecutive differences before calling a list an AP. One matching pair proves nothing — check at least two or three, since $1,1,2,3,5,\ldots$ can start deceptively regular and then break.
Accepting a fractional or negative value of $n$ . If solving $a_n=$ (given number) gives anything other than a positive integer, that number is simply not a term.
Miscounting "kth term from the end." The 11th term from the end of a 25-term AP is the 15th from the start ( $25-11+1$ ), not the 14th; or use $l-(k-1)d$ directly.
Discarding a second valid root of a quadratic. "How many terms give a sum of 78" or "find the AP given a sum and a product" can have two genuinely valid answers; check both.
Assuming $S_n$ and $a_n$ are interchangeable. $S_n$ is a running total; $a_n$ is a single term. The link is $a_n = S_n - S_{n-1}$ , not equality.

Common trapThe single biggest misconception in this chapter is treating the position number n and the number of common-difference jumps as the same thing

Students who have just met the general form $a, a+d, a+2d, a+3d, \ldots$ often assume, without checking, that the term in position $n$ involves exactly $n$ copies of $d$ — writing $a_n = a+nd$ . But $a_1=a$ itself uses zero jumps of $d$ , $a_2$ uses one jump, and in general position $n$ uses exactly $(n-1)$ jumps, one fewer than its position. This single miscount silently corrupts every formula in the chapter at once — the nth term shifts by one full $d$ , the sum formula's bracket $2a+(n-1)d$ goes wrong, and "kth term from the end" calculations go astray the same way. The reliable habit: write out $a_1, a_2, a_3$ explicitly and count the jumps by hand before trusting any formula — $a_1$ has $0$ jumps, $a_2$ has $1$ , $a_3$ has $2$ , and $(n-1)$ jumps for position $n$ becomes impossible to forget.

You should be able to…

State the definition of an AP and identify $a$ and $d$ from a list of terms $a_1, a_2, a_3, \ldots$ .
Test whether an arbitrary list of numbers is an AP by checking $a_{k+1}-a_k$ is constant across several pairs.
Write and apply the nth term formula $a_n=a+(n-1)d$ to find any specific term.
Solve $a_n=$ (given value) for $n$ to check membership, accepting only positive-integer solutions.
Determine an unknown AP from two given terms by solving two simultaneous linear equations in $a$ and $d$ .
Find a term counted from the end of a finite AP using its last term $l$ , without first finding the total number of terms.
Apply $S_n=\tfrac{n}{2}[2a+(n-1)d]$ , and switch to $S_n=\tfrac{n}{2}(a+l)$ when the last term is known directly.
Solve $S_n=$ (given value) for $n$ , recognising the resulting quadratic may have two valid positive-integer roots.
Translate a real-life situation of steady increase or decrease — salary, penalty, savings, production — into an AP.
Use the arithmetic mean relation $b=(a+c)/2$ for three consecutive terms of an AP.

Ask the tutor: Want to work through a mixed batch of "find the term," "find the sum," and "find n" problems back to back until spotting which formula to reach for becomes automatic?

Can the common difference be negative or even zero? I thought a progression is supposed to keep increasing.

Yes to both. The common difference $d$ can be positive (an increasing AP, like $6,9,12,\ldots$ ), negative (a decreasing AP, like $21,18,15,\ldots$ ), or even zero, which gives a constant list like $2,2,2,2,\ldots$ that is technically still an AP. "Progression" here just means the same fixed number is added at every step — it says nothing about the direction that pushes the terms. Always check the sign of $d$ from actual subtraction rather than assuming the list must climb.

Why must d always be found as the later term minus the earlier term, and not the other way round?

Because the sign of $d$ is not bookkeeping — it tells you whether the AP is rising or falling, and every later formula depends on getting that sign right. Subtract in the wrong order even once, say $6-9$ instead of $9-6$ for the AP $6,9,12,\ldots$ , and $d$ comes out $-3$ instead of $3$ , sending every predicted term the wrong way. The safe rule: pick any term, subtract the one immediately before it, always in that order, even when the later term looks smaller.

When I solve for n and get two positive whole-number answers, like 4 and 13, which one is the real answer?

Sometimes both are genuinely correct at once. This happens when the AP has a positive first term and a negative common difference: the terms start positive, cross through negative values, and a block in the middle can sum to exactly zero — which is why the sum reaches the same target twice, at $n=4$ and again at $n=13$ . Substitute both values back into the sum formula to confirm they both work, and report both as valid unless the question's wording (for example, "the smallest such n") tells you to pick one.

What is the difference between the two sum formulas, and how do I know which one to use?

$S_n = \dfrac{n}{2}[2a+(n-1)d]$ and $S_n=\dfrac{n}{2}(a+l)$ are the same formula written two ways, since $a_n=l$ for a finite AP. Use the first form when you know $a$ and $d$ but not the last term directly. Use the second, shorter form whenever $a$ and $l$ are both given outright — it saves finding $d$ at all.