The third and ninth term of an A.P. are 4 and –8 respectively. (i) Which term of the A.P. is zero ? (ii) Find the value of n if S_n = –36.

MathsCBSE Class 10 · Basic · Board exam 2026, set 430/4/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2026: question 32

Set in the CBSE Class 10 board exam of 2026, paper 430/4/1, section D, question 32, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32 · 5 marks · Long answerArithmetic Progressions

The third and ninth term of an A.P. are $4$ and $-8$ respectively.

(i)
Which term of the A.P. is zero ?
0 marks
(ii)
Find the value of $n$ if $S_{n} = -36$ .
0 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 430/4/1.

Hints

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

Hint 1

Write the 3rd and 9th terms using $a_{n} = a + (n - 1) d$ .

Hint 2

Solve the two equations to get a and d, then set $a_{n} = 0$ .

Hint 3

For (ii), use $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ and solve the resulting quadratic; n must be positive.

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.

(i) $a_{3} = a + 2 d = 4$ ...(1)
½ mark
$a_{9} = a + 8 d = - 8$ ...(2)
½ mark
Subtracting (1) from (2): $6 d = - 12 \Rightarrow d = - 2$ ; then $a = 4 - 2 (- 2) = 8$ .
1 mark
$a_{n} = 0 \Rightarrow 8 + (n - 1) (- 2) = 0 \Rightarrow n - 1 = 4 \Rightarrow n = 5$ . So the 5th term is zero.
1 mark
(ii) $S_{n} = \frac{n}{2} [2 a + (n - 1) d] = - 36 \Rightarrow \frac{n}{2} [16 + (n - 1) (- 2)] = - 36$
1 mark
$\Rightarrow \frac{n}{2} (18 - 2 n) = - 36 \Rightarrow 9 n - n^{2} = - 36 \Rightarrow n^{2} - 9 n - 36 = 0$
½ mark
$\Rightarrow (n - 12) (n + 3) = 0 \Rightarrow n = 12$ or $n = - 3$ . Since n is a natural number, reject $n = - 3$ ; hence $n = 12$ .
½ mark

Answer: (i) 5th term; (ii) n = 12

Where the marks are

Writes $a + 2 d = 4$ (½ mark)
Writes $a + 8 d = - 8$ (½ mark)
Solves to get $d = - 2$ , $a = 8$ (½ + ½ marks)
Gets $n = 5$ from $a_{n} = 0$ (1 mark)
Correct $S_{n}$ equation (1 mark)
Forms $n^{2} - 9 n - 36 = 0$ (½ mark)
Factorises and takes $n = 12$ , rejecting $- 3$ (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 430/4/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2026: question 32

Hints

Step-by-step solution

Revise the chapter

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