Answer the following. (a) Find the A.P. whose third term is 16 and seventh term exceeds the fifth term by 12. Also, find the sum of first 29 terms of the A.P. (b) Find the sum of first 20 terms of an A.P. whose n^th term is given by a_n = 5 + 2n. Can 52 be a term of this A.P.?

MathsCBSE Class 10 · Basic · Board exam 2025, set 430/6/1 · 3 marks · Short answer

CBSE Class 10 Mathematics Basic 2025: question 31(a)

Set in the CBSE Class 10 board exam of 2025, paper 430/6/1, section C, question 31(a), for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q31(a) · 3 marks · Short answerArithmetic Progressions

Answer the following.

(a)
Find the A.P. whose third term is 16 and seventh term exceeds the fifth term by 12. Also, find the sum of first 29 terms of the A.P.
0 marks
(b)
Find the sum of first 20 terms of an A.P. whose $n^{t h}$ term is given by $a_{n} = 5 + 2 n$ . Can 52 be a term of this A.P.?
0 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 430/6/1.

Hints

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

Hint 1

Write $a_{3}$ , $a_{5}$ and $a_{7}$ using $a_{n} = a + (n - 1) d$ .

Hint 2

The second condition gives d directly; then use $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ with n = 29.

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.

Let first term = $a$ , common difference = $d$ . Third term: $a + 2 d = 16$ …(i)
½ mark
$a_{7} = a_{5} + 12 \Rightarrow a + 6 d = a + 4 d + 12$ …(ii)
½ mark
From (ii), $2 d = 12 \Rightarrow d = 6$ ; from (i), $a = 16 - 12 = 4$ .
½ mark
The A.P. is 4, 10, 16, 22, …
½ mark
$S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow S_{29} = \frac{29}{2} [8 + 28 \times 6]$
½ mark
$= \frac{29}{2} \times 176 = 29 \times 88 = 2552$
½ mark

Answer: A.P.: 4, 10, 16, …; $S_{29} = 2552$

Where the marks are

Forms $a + 2 d = 16$ (½ mark)
Forms $a + 6 d = a + 4 d + 12$ (½ mark)
Solves to get $d = 6$ , $a = 4$ (½ mark)
Writes the A.P. 4, 10, 16, … (½ mark)
Correct substitution in $S_{29}$ formula (½ mark)
$S_{29} = 2552$ (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 paper2025 board paper, set 430/6/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2025: question 31(a)

Hints

Step-by-step solution

Revise the chapter

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