In a class, the teacher asks every student to write an example of A.P. Two boys Aryan and Roshan writes the progression as −5, −2, 1, 4, …… and 187, 184, 181, …. respectively. Now the teacher asks his various students the following questions on progression. Help the students to find answers for the following: i. Find the sum of the common difference of two progressions. ii. Find the 34th term of progression written by Roshan. iii. (A) Find the sum of first 10 terms of the progression written by Aryan. iii. (B) Which term of the progressions will have the same value?

MathsCBSE Class 10 · Standard · Sample paper, session 2025-26 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard sample paper 2025-26: question 36

From the official CBSE sample question paper for 2025-26, section E, question 36, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q36 · 4 marks · Case-basedCompetency-basedArithmetic Progressions

In a class, the teacher asks every student to write an example of A.P. Two boys Aryan and Roshan writes the progression as −5, −2, 1, 4, …… and 187, 184, 181, …. respectively. Now the teacher asks his various students the following questions on progression.

Help the students to find answers for the following:

i.
Find the sum of the common difference of two progressions.
1 mark
ii.
Find the 34th term of progression written by Roshan.
1 mark
iii.
(A) Find the sum of first 10 terms of the progression written by Aryan.
2 marks
OR
iii.
(B) Which term of the progressions will have the same value?
2 marks

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

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

Hint 1

The common difference is any term minus the term before it.

Hint 2

Use $a_{n} = a + (n - 1) d$ and $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ .

Hint 3

For (iii)(b), set the general terms of the two APs equal and solve for $n$ .

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) Aryan's AP: $d_{1} = - 2 - (- 5) = 3$ .
Roshan's AP: $d_{2} = 184 - 187 = - 3$ .
Sum of the common differences $= 3 + (- 3) = 0$ .
1 mark
(ii) For Roshan's AP, $a = 187$ and $d = - 3$ .
$t_{34} = a + 33 d = 187 + 33 (- 3) = 187 - 99 = 88$
1 mark
(iii)(a) For Aryan's AP, $a = - 5$ , $d = 3$ , $n = 10$ .
$S_{10} = \frac{n}{2} [2 a + (n - 1) d] = \frac{10}{2} [2 (- 5) + 9 (3)]$
1 mark
(iii)(a) $S_{10} = 5 (- 10 + 27) = 5 \times 17 = 85$
1 mark
(iii)(b) Let the $n$ th terms be equal:
$- 5 + (n - 1) 3 = 187 + (n - 1) (- 3)$
1 mark
(iii)(b) $6 (n - 1) = 192$
$\Rightarrow n - 1 = 32$
$\Rightarrow n = 33$
So the 33rd terms are equal. Check: both equal $- 5 + 96 = 187 - 96 = 91$ .
1 mark

Answer: (i) 0; (ii) 88; (iii)(a) 85, (iii)(b) 33rd term (value 91)

Where the marks are

(i) Finds $d = 3$ and $d = - 3$ , sum 0 (1 mark)
(ii) $t_{34} = 88$ (1 mark)
(iii)(a) Correct substitution in the $S_{n}$ formula (1 mark); $S_{10} = 85$ (1 mark)
(iii)(b) Equates the two $n$ th terms (1 mark); $n = 33$ (1 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 paperSample paper 2025-26 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard sample paper 2025-26: question 36

Hints

Step-by-step solution

Revise the chapter

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