The sum of first n terms of an A.P. is 2n² + 13n. Find its n^th term and hence 10^th term.

MathsCBSE Class 10 · Standard · Board exam 2026, set 30/5/1 · 3 marks · Short answer

CBSE Class 10 Mathematics Standard 2026: question 27(b)

Set in the CBSE Class 10 board exam of 2026, paper 30/5/1, section C, question 27(b), for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q27(b) · 3 marks · Short answerArithmetic Progressions

The sum of first n terms of an A.P. is $2 n^{2} + 13 n$ . Find its $n^{t h}$ term and hence $1 0^{t h}$ term.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 30/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

The sum of the first one term is just the first term.

Hint 2

Find $S_{2}$ and subtract $a_{1}$ to get $a_{2}$ .

Hint 3

Use $a_{n} = a + (n - 1) d$ (or $a_{n} = S_{n} - S_{n - 1}$ ).

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

$S_{n} = 2 n^{2} + 13 n$ . $S_{1} = a_{1} = 2 + 13 = 15$
½ mark
$S_{2} = a_{1} + a_{2} = 8 + 26 = 34 \Rightarrow a_{2} = 34 - 15 = 19$
½ mark
$d = a_{2} - a_{1} = 19 - 15 = 4$
½ mark
$a_{n} = a + (n - 1) d = 15 + (n - 1) \times 4 = 4 n + 11$
1 mark
$a_{10} = 4 \times 10 + 11 = 51$
½ mark

Answer: $a_{n} = 4 n + 11$ ; $a_{10} = 51$

Where the marks are

$a_{1} = 15$ (½ mark)
$a_{2} = 19$ (½ mark)
$d = 4$ (½ mark)
$a_{n} = 4 n + 11$ (1 mark)
$a_{10} = 51$ (½ mark)

Revise the chapter

Chapter: Arithmetic Progressions (notes, formulas, where marks are lost and every past question)
Topic: Finding nth Term from the Sum of n Terms

Practise this chapterBoard-style questions from Arithmetic Progressions, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 30/5/1 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard 2026: question 27(b)

Hints

Step-by-step solution

Revise the chapter

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