MathsICSE Class 10 · Arithmetic and Geometric Progression

Sum of n Terms of an A.P.: Formulas, Finding n and Solved Questions (ICSE Class 10)

Q: What is the formula for the sum of n terms of an A.P.?

Sn = n/2 [2a + (n − 1)d], where a is the first term and d the common difference. If the last term l is known, Sn = n/2 (a + l) is quicker.

Q: Why do I sometimes get two values of n?

Finding n from a sum gives a quadratic. Reject negative or fractional roots. If both roots are positive whole numbers (possible when d is negative), both are valid because the extra terms add up to zero.

Q: How do I find the nth term from the sum formula?

Use Tn = Sn − Sn−1, and T1 = S1. For example, if Sn = 3n² − n, then Tn = 6n − 4.

Updated 15 October 2026 · Fact-checked

The sum of the first n terms of an A.P. is Sn = n/2 [2a + (n − 1)d], or Sn = n/2 (a + l) when the last term l is known. To find n from a sum, solve the resulting quadratic and keep positive whole-number roots. Also, Tn = Sn − Sn−1.

Each column of the forwards and reversed sums totals a + l, and there are n columns, so 2Sn = n(a + l) and Sn = n/2 (a + l).

Illustration drawn for QuestionBuddy.

Understand Sum of n Terms of an A.P.

Where the formula comes from. Write the sum forwards and backwards:

$S_{n} = a + (a + d) + \dots + l$ and $S_{n} = l + (l - d) + \dots + a$ .

Adding pairs, every column totals $a + l$ , and there are $n$ columns, so $2 S_{n} = n (a + l)$ :

$S_{n} = \frac{n}{2} (a + l) = \frac{n}{2} [2 a + (n - 1) d]$ , since $l = a + (n - 1) d$ .

Which form to use.

Know $a$ , $d$ and $n$ : use $\frac{n}{2} [2 a + (n - 1) d]$ .
Know the first and last terms: use $\frac{n}{2} (a + l)$ . In 2025 Q4(ii) ( $a = 5$ , $l = 50$ , $S = 440$ ) this gives $\frac{n}{2} \times 55 = 440$ , so $n = 16$ at once; the examiners noted that candidates who used the longer formula made more slips.

Finding n from a sum. Substituting into $\frac{n}{2} [2 a + (n - 1) d] = S$ gives a quadratic in $n$ . Keep only positive whole-number roots. If both roots are valid (this can happen when $d < 0$ and some terms are negative), both answers are correct: the extra terms add up to zero.

**From $T_{n}$ to $S_{n}$ and back.** If $T_{n}$ is given as a formula, find $a = T_{1}$ and $d = T_{2} - T_{1}$ and then use $S_{n}$ ; in 2024 Q6(ii), $T_{n} = 6 (7 - n)$ gave $S_{25} = - 900$ , and the examiners reported candidates getting $+ 900$ by sign errors. If $S_{n}$ is given as a formula, then $T_{1} = S_{1}$ and $T_{n} = S_{n} - S_{n - 1}$ .

Useful special sums. $1 + 2 + \dots + n = \frac{n ( n + 1 )}{2}$ ; $1 + 3 + 5 + \dots$ ( $n$ odd numbers) $= n^{2}$ .

Do not confuse the last term $l$ with the sum $S$ . Read carefully which number is which: the 2025 report noted candidates who took 50 as $n$ and 440 as the last term.

Sum of n Terms of an A.P.: formulas to remember

Sum with a and d: $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$; The standard form.
Sum with first and last terms: $S_{n} = \frac{n}{2} (a + l)$; Quicker when $l$ is known.
Term from sums: $T_{n} = S_{n} - S_{n - 1}, T_{1} = S_{1}$; Use when $S_{n}$ is a formula in $n$ .
First n natural numbers: $1 + 2 + \dots + n = \frac{n ( n + 1 )}{2}$; An A.P. with $a = d = 1$ .
First n odd numbers: $1 + 3 + 5 + \dots + (2 n - 1) = n^{2}$; An A.P. with $a = 1$ , $d = 2$ .

How to solve Sum of n Terms of an A.P. questions

Name what you know (a, d, l, n or S), pick the matching form, then solve.

1List the knowns: $a$ , $d$ , $n$ , $l$ , $S_{n}$ .
2If the last term is known, prefer $S_{n} = \frac{n}{2} (a + l)$ .
3If $n$ is unknown, substitute everything else and form a quadratic in $n$ ; factorise or use the formula.
4Reject negative or fractional values of $n$ ; explain any second valid value.
5If $d$ is still needed, use $l = a + (n - 1) d$ .
6Check by a quick estimate: the sum is $n$ times the average of first and last terms.

Sum of n Terms of an A.P. 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.

Sum of an A.P. for a total count

Suppose one section of a stadium has 25 rows; the front row has 30 seats and each row behind has 4 more seats than the one in front.

How many seats are in the last row and in the whole section?

Try it first, then show the solutionSolution

$a = 30$ , $d = 4$ , $n = 25$ : last row $l = 30 + 24 \times 4 = 126$ .
$S_{25} = \frac{25}{2} (30 + 126) = \frac{25}{2} \times 156 = 1950$ .

Answer: 126 seats in the last row; 1950 seats in all.

Memory hook: Total = number of rows × average of the first and last rows.

Sum of an A.P. of salaries

Imagine a job paying ₹30,000 a month in the first year, with the monthly pay rising by ₹2,000 at the start of each new year.

Find the total earned over the first 10 years.

Try it first, then show the solutionSolution

Monthly pay in years 1 to 10: 30000, 32000, …, an A.P. with last term $30000 + 9 \times 2000 = 48000$ .
Sum of the 10 monthly rates: $\frac{10}{2} (30000 + 48000) = 390000$ .
Each rate is paid for 12 months: total $= 12 \times 390000 = 4680000$ .

Answer: ₹46,80,000

Memory hook: Add the A.P. first, then scale by how often each term repeats.

Tricks to remember Sum of n Terms of an A.P.

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

Average times count
$S_{n} = n \times \frac{a + l}{2}$ : how many terms times the middle value.
Use it when: Quick sums and sanity checks.
Gauss's pairing
Pair the first with the last, the second with the second-last: each pair adds to $a + l$ , and there are $\frac{n}{2}$ pairs.
Use it when: Rebuilding the formula if you forget it.
S minus S gives T
The $n$ th term is what the $n$ th step added: $T_{n} = S_{n} - S_{n - 1}$ .
Use it when: Questions that give $S_{n}$ as a formula.

Sum of n Terms of an A.P.: solved examples

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

Example 1: A direct sum · 1 mark · MCQBoard level

The sum of the first 10 terms of the A.P. 2, 7, 12, … is:
(a) 245
(b) 250
(c) 235
(d) 490

Try it first, then show the solutionSolution

$a = 2$ , $d = 5$ , $n = 10$ .
$S_{10} = 5 [4 + 9 \times 5] = 5 \times 49 = 245$ .

Answer: (a) 245

Example 2: First term, last term and sum · 3 marks · Short answerBoard level

The first term of an A.P. is 8, its last term is 62 and the sum of all its terms is 350. Find the number of terms and the common difference.

Try it first, then show the solutionSolution

$S_{n} = \frac{n}{2} (a + l)$ : $\frac{n}{2} (8 + 62) = 350 \Rightarrow 35 n = 350 \Rightarrow n = 10$ .
$l = a + (n - 1) d$ : $62 = 8 + 9 d \Rightarrow d = 6$ .

Answer: 10 terms; $d = 6$ .

Example 3: Sum when Tn is given · 3 marks · Short answerHarder board level

The $n$ th term of an A.P. is $T_{n} = 2 (11 - n)$ . Find (a) $a$ and $d$ (b) the sum of its first 30 terms.

Try it first, then show the solutionSolution

(a) $T_{1} = 20$ , $T_{2} = 18$ , so $a = 20$ , $d = - 2$ .
(b) $S_{30} = 15 [40 + 29 (- 2)] = 15 (40 - 58) = 15 (- 18) = - 270$ .

Answer: (a) $a = 20$ , $d = - 2$ (b) $- 270$

Example 4: How many terms? · 4 marks · Short answerHarder board level

How many terms of the A.P. 3, 7, 11, … must be taken to make a sum of 406?

Try it first, then show the solutionSolution

$\frac{n}{2} [6 + (n - 1) 4] = 406 \Rightarrow \frac{n}{2} (4 n + 2) = 406 \Rightarrow n (2 n + 1) = 406$ .
$2 n^{2} + n - 406 = 0$ . Discriminant $= 1 + 3248 = 3249 = 5 7^{2}$ .
$n = \frac{- 1 \pm 57}{4}$ : $n = 14$ or $n = - 14.5$ (rejected).

Answer: 14 terms.

Example 5: Two partial sums · 4 marks · Short answerHOTS / competency

The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the A.P.

Try it first, then show the solutionSolution

$S_{7} = \frac{7}{2} (2 a + 6 d) = 63 \Rightarrow a + 3 d = 9$ … (1)
$S_{14} = 63 + 161 = 224$ : $7 (2 a + 13 d) = 224 \Rightarrow 2 a + 13 d = 32$ … (2)
From (1), $2 a + 6 d = 18$ ; subtracting from (2): $7 d = 14$ , $d = 2$ , $a = 3$ .

Answer: 3, 5, 7, 9, …

Example 6: Case study: weekly savings · 4 marks · Case-basedHOTS / competency

Arjun saves ₹200 in week 1 and ₹25 more each week than the week before.
(a) How much does he save in week 12?
(b) What has he saved in total after 12 weeks?
(c) After how many weeks will his total be ₹8,750?

Try it first, then show the solutionSolution

(a) $T_{12} = 200 + 11 \times 25 = 475$ .
(b) $S_{12} = 6 (200 + 475) = 4050$ .
(c) $\frac{n}{2} [400 + 25 (n - 1)] = 8750 \Rightarrow n (375 + 25 n) = 17500 \Rightarrow n^{2} + 15 n - 700 = 0$ .
$(n + 35) (n - 20) = 0$ , so $n = 20$ .

Answer: (a) ₹475 (b) ₹4,050 (c) 20 weeks

Where students lose marks in Sum of n Terms of an A.P.

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

Mistake: Mixing up the last term and the sum
Read which number is $l$ and which is $S_{n}$ .
Wrong working
First term 5, last term 50, sum 440: take $n = 50$ and $T_{n} = 440$ .
Right working
$\frac{n}{2} (5 + 50) = 440 \Rightarrow n = 16$ .
Do this: Write 'a = …, l = …, S = …' before any working.
Usually costs: Reported in 2025 Q4(ii).
Mistake: Sign slips with a negative d
$(n - 1) d$ is negative when $d < 0$ .
Wrong working
$S_{25} = \frac{25}{2} [72 + 24 \times 6] = 2700$ or $+ 900$ .
Right working
$S_{25} = \frac{25}{2} [72 + 24 (- 6)] = - 900$ .
Do this: Put $d$ in brackets with its sign.
Usually costs: Reported in 2024 Q6(ii).
Mistake: Using Tn instead of Sn
The sum adds all terms; $T_{n}$ is just the last.
Wrong working
Sum of 25 terms $= T_{25} = - 108$ .
Right working
Use $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ .
Do this: Underline 'sum' in the question.
Usually costs: All marks for that part (2024 Q6(ii) report).

Exam tips

2025 Q4(ii) (3 marks) expected $S_{n} = \frac{n}{2} (a + l)$ to get $n = 16$ , then $l = a + (n - 1) d$ to get $d = 3$ .
2024 Q6(ii) expected $T_{1} = 36$ , $T_{2} = 30$ , then $S_{25} = - 900$ ; keep the negative sign.
When a quadratic in $n$ appears, show the factorisation and state why you reject a root.
Write the formula in letters before substituting; the method mark often rests on it.

Practice questions on Sum of n Terms of an A.P.

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 20 natural numbers is:
(a) 200
(b) 210
(c) 220
(d) 420
Hint
$\frac{n ( n + 1 )}{2}$ .
Answer
(b) 210
Q2 · 2 marks · Short answerBoard level
Find the sum of the first 12 terms of 5, 9, 13, …
Hint
$a = 5$ , $d = 4$ .
Answer
$6 [10 + 44] = 324$
Q3 · 3 marks · Short answerHarder board level
Find the sum of all odd numbers between 0 and 100.
Hint
1, 3, …, 99 has 50 terms.
Answer
$\frac{50}{2} (1 + 99) = 2500$
Q4 · 3 marks · Short answerHarder board level
Find the sum of all multiples of 6 between 100 and 300.
Hint
102, 108, …, 294.
Answer
$n = \frac{294 - 102}{6} + 1 = 33$ ; $S = \frac{33}{2} (102 + 294) = 6534$ .
Q5 · 3 marks · Short answerHarder board level
In an A.P., the first term is −4, the last term is 146 and the sum is 1846. Find the number of terms and the common difference.
Hint
$\frac{n}{2} (- 4 + 146) = 1846$ .
Answer
$71 n = 1846$ , $n = 26$ ; $146 = - 4 + 25 d$ , $d = 6$ .
Q6 · 4 marks · Short answerHOTS / competency
The sum of the first $n$ terms of an A.P. is $S_{n} = 3 n^{2} - n$ . Find its first term, common difference and 10th term.
Hint
$T_{n} = S_{n} - S_{n - 1}$ .
Answer
$T_{n} = 6 n - 4$ , so $a = 2$ , $d = 6$ , $T_{10} = 56$ .
Q7 · 3 marks · Short answerHarder board level
Show that the sum of the first $n$ odd numbers is $n^{2}$ . How many consecutive odd numbers, starting from 1, add up to 625?
Hint
$a = 1$ , $d = 2$ .
Answer
$S_{n} = \frac{n}{2} [2 + 2 (n - 1)] = n^{2}$ ; $n^{2} = 625$ gives 25 numbers.

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.Write the two formulas for $S_{n}$ .Show answer
$\frac{n}{2} [2 a + (n - 1) d]$ and $\frac{n}{2} (a + l)$ .
2.Find $1 + 2 + \dots + 100$ .Show answer
$\frac{100 \times 101}{2} = 5050$
3.If $S_{n}$ is known as a formula, how do you get $T_{n}$ ?Show answer
$T_{n} = S_{n} - S_{n - 1}$ .
4.What is the sum of the first 10 odd numbers?Show answer
$1 0^{2} = 100$
5.If $a = 3$ , $l = 27$ and $n = 9$ , find $S_{n}$ .Show answer
$\frac{9}{2} \times 30 = 135$

Sum of n Terms of an A.P.: frequently asked questions

What is the formula for the sum of n terms of an A.P.?

Sn = n/2 [2a + (n − 1)d], where a is the first term and d the common difference. If the last term l is known, Sn = n/2 (a + l) is quicker.

Why do I sometimes get two values of n?

Finding n from a sum gives a quadratic. Reject negative or fractional roots. If both roots are positive whole numbers (possible when d is negative), both are valid because the extra terms add up to zero.

How do I find the nth term from the sum formula?

Use Tn = Sn − Sn−1, and T1 = S1. For example, if Sn = 3n² − n, then Tn = 6n − 4.