MathsCBSE Class 10 · Arithmetic Progressions

Finding the nth Term from the Sum of n Terms of an AP

Updated 13 October 2026 · Fact-checked

When the sum of the first n terms of an AP is given as an expression in n, you find the nth term by subtracting the sum of the first n minus 1 terms from it. The first term is the sum of one term. Board questions ask for the nth term, the common difference, or a particular term.

S5 and S4 share a1 to a4, so subtracting leaves only the last term: a_n = S_n − S_(n−1), for n ≥ 2, with a1 = S1.

Illustration drawn for QuestionBuddy.

Understand Finding nth Term from the Sum of n Terms

Sometimes a question does not give you the AP. It gives you a rule for the running total instead, such as $S_{n} = 5 n^{2} - 3 n$ . Your job is to recover the terms from the total.

The idea is simple. $S_{n}$ is the sum of the first $n$ terms, and $S_{n - 1}$ is the sum of the first $n - 1$ terms. Both sums contain $a_{1}, a_{2}, \dots, a_{n - 1}$ . The only term in $S_{n}$ that is missing from $S_{n - 1}$ is $a_{n}$ . So subtracting gives $a_{n} = S_{n} - S_{n - 1}$ . For example, $S_{5} - S_{4} = a_{5}$ .

This works for $n \geq 2$ , because $S_{0}$ is not part of the list. For the first term, use $a_{1} = S_{1}$ directly: the sum of one term is that term. Always check that your formula for $a_{n}$ gives $S_{1}$ when you put $n = 1$ .

This links to the sum formula you derived earlier. Expanding $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ gives $S_{n} = \frac{d}{2} n^{2} + (a - \frac{d}{2}) n$ . So the sum of an AP is always of the form $p n^{2} + q n$ , a quadratic in $n$ with no constant term. Then $a_{n} = 2 p n + (q - p)$ and $d = 2 p$ . The common difference is twice the coefficient of $n^{2}$ , not equal to it.

If a given $S_{n}$ has a constant term, like $n^{2} + 2 n + 3$ , the list is not an AP. The rule $a_{n} = S_{n} - S_{n - 1}$ still gives the terms from the second onwards, but $a_{1} = S_{1}$ no longer fits the same pattern.

Once you have $a_{n}$ , everything from the earlier topics opens up: find any term, find which term equals a number, or find the sum of a block such as the 11th to 20th terms using $S_{20} - S_{10}$ .

Finding nth Term from the Sum of n Terms: formulas to remember

nth term from the sum: $a_{n} = S_{n} - S_{n - 1}, n \geq 2$; Replace every $n$ by $(n - 1)$ inside brackets when you write $S_{n - 1}$ .
First term from the sum: $a_{1} = S_{1}$; Put $n = 1$ in the sum expression.
Common difference from sums: $d = a_{2} - a_{1} = S_{2} - 2 S_{1}$; Since $a_{2} = S_{2} - S_{1}$ and $a_{1} = S_{1}$ .
Shape of S_n for an AP: $S_{n} = p n^{2} + q n \Rightarrow a_{n} = 2 p n + (q - p), d = 2 p$; No constant term. The common difference is twice the coefficient of $n^{2}$ .
Sum of a block of terms: $a_{m + 1} + \dots + a_{k} = S_{k} - S_{m}$; For the 11th to 20th terms use $S_{20} - S_{10}$ , not $S_{20} - S_{11}$ .

How to solve Finding nth Term from the Sum of n Terms questions

This method works for any question where $S_{n}$ is given as an expression in $n$ .

1Write down $S_{n}$ and find $S_{1}$ . This is $a_{1}$ .
2Write $S_{n - 1}$ by replacing every $n$ with $(n - 1)$ , using brackets, and expand it fully.
3Subtract: $a_{n} = S_{n} - S_{n - 1}$ . Simplify to the form $A n + B$ .
4Check: put $n = 1$ in $a_{n}$ . It must equal $S_{1}$ . If it does not, the list is not an AP.
5Read off $d$ as the coefficient of $n$ in $a_{n}$ (or as twice the coefficient of $n^{2}$ in $S_{n}$ ).
6Use $a_{n}$ to answer the actual question: a particular term, which term equals a value, or a block sum $S_{k} - S_{m}$ .
7Write the final answer clearly, with units in a word problem.

Finding nth Term from the Sum of n Terms in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Finding the nth term when the sum of n terms is given

Suppose you track your mobile data during a two-month holiday. You notice that the total data you have used by the end of day $n$ is always $S_{n} = 10 n^{2} + 190 n$ megabytes.

How much data do you use on day 1, and on day 15 alone? On which day do you first use more than 600 MB in a single day?

Try it first, then show the solutionSolution

$a_{1} = S_{1} = 10 + 190 = 200$ MB.
$S_{n - 1} = 10 (n - 1)^{2} + 190 (n - 1) = 10 n^{2} + 170 n - 180$ .
$a_{n} = S_{n} - S_{n - 1} = 20 n + 180$ . Check: $a_{1} = 200$ ✓.
$a_{15} = 20 (15) + 180 = 480$ MB.
$20 n + 180 > 600 \Rightarrow n > 21$ .
Day 21 uses exactly 600 MB, so the first day above 600 MB is day 22.

Answer: 200 MB on day 1, 480 MB on day 15; day 22 is the first day with more than 600 MB.

Memory hook: Total till today minus total till yesterday equals today's data.

Finding the nth term when the sum of n terms is given

Imagine your elder sister saves for a laptop. Her total savings at the end of month $n$ are $S_{n} = 25 n^{2} + 75 n$ rupees.

Find how much she saves in the 12th month, and in which month she saves exactly ₹1,000.

Try it first, then show the solutionSolution

$S_{n - 1} = 25 (n - 1)^{2} + 75 (n - 1) = 25 n^{2} + 25 n - 50$ .
$a_{n} = S_{n} - S_{n - 1} = 50 n + 50$ . Check: $a_{1} = 100 = S_{1}$ ✓.
$a_{12} = 50 (12) + 50 = 650$ .
$50 n + 50 = 1000 \Rightarrow n = 19$ .

Answer: ₹650 in the 12th month; she saves ₹1,000 in the 19th month.

Memory hook: The month's saving is the jump in the running total.

Tricks to remember Finding nth Term from the Sum of n Terms

Short memory aids. Say each one aloud once, then use it on a question the same day.

Today = Total till today − Total till yesterday
$a_{n} = S_{n} - S_{n - 1}$ is like finding today's spending from two bank balances.
Use it when: Whenever $S_{n}$ is given and a term is asked.
Double the square
In $S_{n} = p n^{2} + q n$ , $d = 2 p$ . Say: the difference doubles the square's coefficient.
Use it when: Quick check of $d$ in MCQs and assertion-reason.
No constant, it's constant
An AP sum has no constant term; then the difference stays constant.
Use it when: Deciding whether a given $S_{n}$ belongs to an AP.

Finding nth Term from the Sum of n Terms: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Find a, d and a term from S_n · 3 marks · Short answerBoard level

The sum of the first $n$ terms of an AP is $S_{n} = 5 n^{2} - 3 n$ . Find the first term, the common difference, the $n$ th term and the 15th term.

Try it first, then show the solutionSolution

$a_{1} = S_{1} = 5 - 3 = 2$ . (½ mark)
$S_{n - 1} = 5 (n - 1)^{2} - 3 (n - 1) = 5 n^{2} - 10 n + 5 - 3 n + 3 = 5 n^{2} - 13 n + 8$ . (1 mark)
$a_{n} = S_{n} - S_{n - 1} = (5 n^{2} - 3 n) - (5 n^{2} - 13 n + 8) = 10 n - 8$ . (½ mark)
Check: $a_{1} = 10 - 8 = 2 = S_{1}$ ✓. So $d = 10$ . (½ mark)
$a_{15} = 150 - 8 = 142$ . (½ mark)

Answer: $a = 2$ , $d = 10$ , $a_{n} = 10 n - 8$ , $a_{15} = 142$ .

Example 2: Which term, and a block sum · 3 marks · Short answerHarder board level

The sum of the first $n$ terms of an AP is $3 n^{2} + 5 n$ . Which term of the AP is 164? Also find the sum of its 11th to 20th terms.

Try it first, then show the solutionSolution

$S_{n - 1} = 3 (n - 1)^{2} + 5 (n - 1) = 3 n^{2} - n - 2$ . (½ mark)
$a_{n} = (3 n^{2} + 5 n) - (3 n^{2} - n - 2) = 6 n + 2$ ; check $a_{1} = 8 = S_{1}$ ✓. (½ mark)
$6 n + 2 = 164 \Rightarrow n = 27$ , a natural number. (1 mark)
Sum of 11th to 20th terms $= S_{20} - S_{10} = (1200 + 100) - (300 + 50) = 950$ . (1 mark)
Check: $a_{11} = 68$ , $a_{20} = 122$ , $\frac{10}{2} (68 + 122) = 950$ ✓.

Answer: 164 is the 27th term; the sum of the 11th to 20th terms is 950.

Example 3: Unknown constant in S_n · 3 marks · Short answerHarder board level

The sum of the first $n$ terms of an AP is $S_{n} = k n^{2} + 2 n$ . If its 5th term is 47, find $k$ , the common difference and the sum of the first 20 terms.

Try it first, then show the solutionSolution

$a_{5} = S_{5} - S_{4} = (25 k + 10) - (16 k + 8) = 9 k + 2$ . (1 mark)
$9 k + 2 = 47 \Rightarrow k = 5$ . (½ mark)
$S_{n} = 5 n^{2} + 2 n$ , so $d = 2 \times 5 = 10$ ; also $a_{1} = S_{1} = 7$ . (½ mark)
$S_{20} = 5 (400) + 40 = 2040$ . (1 mark)
Check: $a_{20} = 7 + 19 (10) = 197$ , $\frac{20}{2} (7 + 197) = 2040$ ✓.

Answer: $k = 5$ , $d = 10$ , $S_{20} = 2040$ .

Example 4: Long answer: nth term, AP check and number of terms · 5 marks · Long answerHarder board level

The sum of the first $n$ terms of a sequence is $S_{n} = \frac{3 n ^{2} + 7 n}{2}$ . (i) Find the $n$ th term and show that the sequence is an AP. (ii) Find its 20th term. (iii) How many terms must be taken to get a sum of 185?

Try it first, then show the solutionSolution

$S_{n - 1} = \frac{3 ( n - 1 ) ^{2} + 7 ( n - 1 )}{2} = \frac{3 n ^{2} + n - 4}{2}$ . (1 mark)
$a_{n} = S_{n} - S_{n - 1} = \frac{6 n + 4}{2} = 3 n + 2$ . (½ mark)
$a_{1} = S_{1} = \frac{10}{2} = 5$ and $3 (1) + 2 = 5$ ✓. (½ mark)
$a_{n + 1} - a_{n} = 3$ , a constant, so it is an AP with $d = 3$ . (½ mark)
$a_{20} = 62$ . (½ mark)
$\frac{3 n ^{2} + 7 n}{2} = 185 \Rightarrow 3 n^{2} + 7 n - 370 = 0$ . (½ mark)
$(n - 10) (3 n + 37) = 0$ , so $n = 10$ or $n = - \frac{37}{3}$ . (1 mark)
$n$ must be a natural number, so $n = 10$ . (½ mark)

Answer: $a_{n} = 3 n + 2$ , an AP with $d = 3$ ; $a_{20} = 62$ ; 10 terms give a sum of 185.

Example 5: Find the error: is it an AP? · 3 marks · Short answerHOTS / competency

The sum of the first $n$ terms of a sequence is $S_{n} = n^{2} + 2 n + 3$ . A student writes: " $a_{n} = S_{n} - S_{n - 1} = 2 n + 1$ , so the sequence is an AP with $d = 2$ and first term 3." Find the error and decide whether the sequence is an AP.

Try it first, then show the solutionSolution

$a_{n} = S_{n} - S_{n - 1} = 2 n + 1$ is correct only for $n \geq 2$ . (½ mark)
The first term must come from $a_{1} = S_{1} = 1 + 2 + 3 = 6$ , not from $2 (1) + 1 = 3$ . (1 mark)
So the terms are $6, 5, 7, 9, \dots$ . (½ mark)
$a_{2} - a_{1} = - 1$ but $a_{3} - a_{2} = 2$ ; differences are unequal. (½ mark)
Reason: an AP sum has no constant term; here the constant 3 breaks the pattern. (½ mark)

Answer: The student wrongly took $a_{1} = 3$ ; actually $a_{1} = 6$ , so the sequence $6, 5, 7, 9, \dots$ is not an AP.

Where students lose marks in Finding nth Term from the Sum of n Terms

The slips examiners see most often, and what to write instead.

Mistake: Dividing $S_{n}$ by $n$ to get $a_{n}$
$\frac{S _{n}}{n}$ is the average of the first $n$ terms, not the last term.
Wrong working
$S_{n} = 5 n^{2} - 3 n \Rightarrow a_{n} = 5 n - 3$
Right working
$a_{n} = S_{n} - S_{n - 1} = 10 n - 8$
Do this: Always subtract consecutive sums.
Usually costs: Usually all marks for that part.
Mistake: Dropping brackets in $S_{n - 1}$
A minus sign before $S_{n - 1}$ must change every sign inside.
Wrong working
$a_{n} = 3 n^{2} + 5 n - 3 n^{2} - n - 2 + \dots$ written as $3 n^{2} + 5 n - 3 n^{2} - n + 2 = 4 n + 2$
Right working
$(3 n^{2} + 5 n) - (3 n^{2} - n - 2) = 6 n + 2$
Do this: Write $S_{n - 1}$ in full brackets before subtracting.
Usually costs: 1 to 2 marks, and every later part goes wrong.
Mistake: Taking $d$ equal to the coefficient of $n^{2}$
$S_{n} = \frac{d}{2} n^{2} + \dots$ , so the coefficient is half of $d$ .
Wrong working
$S_{n} = 2 n^{2} + 5 n \Rightarrow d = 2$
Right working
$d = 2 \times 2 = 4$ , since $a_{2} - a_{1} = 11 - 7$
Do this: Confirm $d$ using $a_{2} - a_{1}$ .
Usually costs: The full 1 mark in an MCQ.
Mistake: Using $S_{20} - S_{11}$ for the 11th to 20th terms
$S_{11}$ already contains the 11th term, so it gets removed.
Wrong working
$S_{20} - S_{11}$
Right working
$S_{20} - S_{10}$
Do this: Subtract the sum up to the term just before the block.
Usually costs: 1 mark or more.

Exam tips

Show the line $a_{n} = S_{n} - S_{n - 1}$ and the expanded $S_{n - 1}$ ; examiners give step marks for both.
Check $a_{1} = S_{1}$ in one line. It catches most algebra slips.
In MCQs on $d$ , use $d = 2 \times$ (coefficient of $n^{2}$ ), then confirm with $S_{2} - 2 S_{1}$ .
If a part asks only for one term, $a_{k} = S_{k} - S_{k - 1}$ with numbers is faster than finding $a_{n}$ .
When a sum is given and $n$ is asked, reject non-natural roots with a stated reason.

Practice questions on Finding nth Term from the Sum of n Terms

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The sum of the first $n$ terms of an AP is $2 n^{2} + 5 n$ . Its common difference is
(a) 4
(b) 2
(c) 7
(d) 9
Hint
$d$ is twice the coefficient of $n^{2}$ , or find $a_{2} - a_{1}$ .
Answer
(a) 4. $a_{1} = S_{1} = 7$ , $a_{2} = S_{2} - S_{1} = 18 - 7 = 11$ , so $d = 4$ .
Q2 · 3 marks · Short answerHarder board level
The sum of the first $n$ terms of an AP is $6 n - n^{2}$ . Find its $n$ th term and find which term is $- 9$ .
Hint
Expand $6 (n - 1) - (n - 1)^{2}$ carefully with brackets.
Answer
$S_{n - 1} = - n^{2} + 8 n - 7$ , so $a_{n} = 7 - 2 n$ (check $a_{1} = 5 = S_{1}$ ). $7 - 2 n = - 9 \Rightarrow n = 8$ . The 8th term is $- 9$ .
Q3 · 3 marks · Short answerHarder board level
For an AP, $S_{7} = 49$ and $S_{17} = 289$ . Find $S_{n}$ and the $n$ th term.
Hint
Take $S_{n} = p n^{2} + q n$ and form two equations.
Answer
$49 p + 7 q = 49 \Rightarrow 7 p + q = 7$ ; $289 p + 17 q = 289 \Rightarrow 17 p + q = 17$ . So $p = 1$ , $q = 0$ , $S_{n} = n^{2}$ , and $a_{n} = 2 n - 1$ .
Q4 · 3 marks · Short answerHarder board level
The sum of the first $n$ terms of an AP is $n^{2} + 4 n$ . Which term of the AP is three times its 4th term? Also find the sum of the terms from the 5th to the 15th.
Hint
Find $a_{n}$ first; the block sum is $S_{15} - S_{4}$ .
Answer
$a_{n} = 2 n + 3$ , $a_{4} = 11$ . $2 n + 3 = 33 \Rightarrow n = 15$ , the 15th term. $S_{15} - S_{4} = 285 - 32 = 253$ (check: 11 terms from 13 to 33, $\frac{11}{2} (46) = 253$ ).
Q5 · 4 marks · Case-basedHOTS / competency
Suppose a small factory raises its output by a fixed number of units every day. Its total output by the end of day $n$ is $S_{n} = 3 n^{2} + 37 n$ units.
(i) How many units are made on day 1? (1 mark)
(ii) By how many units does daily output rise each day? (1 mark)
(iii) Find the output on day 20 and the total output from day 11 to day 20. (2 marks)
Hint
$a_{1} = S_{1}$ ; the daily rise is $d = 2 p$ ; the block total is $S_{20} - S_{10}$ .
Answer
(i) $S_{1} = 40$ units. (ii) $a_{n} = 6 n + 34$ , so 6 units. (iii) $a_{20} = 154$ units; $S_{20} - S_{10} = 1940 - 670 = 1270$ units.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Formula for $a_{n}$ from $S_{n}$ ?Show answer
$a_{n} = S_{n} - S_{n - 1}$ for $n \geq 2$ .
2.How do you get the first term?Show answer
$a_{1} = S_{1}$ .
3.If $S_{n} = p n^{2} + q n$ , what is $d$ ?Show answer
$d = 2 p$ .
4.If $S_{n} = 4 n^{2} - n$ , what is $a_{3}$ ?Show answer
$S_{3} - S_{2} = 33 - 14 = 19$ .
5.Sum of the 6th to 10th terms in terms of sums?Show answer
$S_{10} - S_{5}$ .

Finding nth Term from the Sum of n Terms: frequently asked questions

Why does subtracting S of n minus 1 from S of n give the nth term?

Both sums contain the first n minus 1 terms. The only extra term in the sum of n terms is the nth term, so the difference is exactly that term.

Why can I not use the formula for the first term?

The sum of zero terms is not part of the list, so the rule works from the second term. For the first term just put n equal to 1 in the sum.

How can I tell if a given sum formula belongs to an AP?

For an AP the sum is a quadratic in n with no constant term. If there is a constant term, the first term does not fit the pattern and the list is not an AP.

Is the optional Exercise 5.4 tested in the board exam?

No. The NCERT book marks Exercise 5.4 as optional and not assessed, so focus on Exercises 5.2 and 5.3.