Q) 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. Find the sum of first 10 terms of the progression written by Aryan.
iv. Which term of the progressions will have the same value?

Q36 – Sample Question Paper – Set 1 – Maths Standard – CBSE 2026

Ans:

VIDEO SOLUTION

STEP BY STEP SOLUTION

Given that Aryan’s AP is −5, −2, 1, 4, ……

Roshan’s AP is 187, 184, 181, ….

i. Sum of the common difference of two progressions:

We know that common difference of an AP is the difference between every two consecutive terms, let’s calculate common differences of both APs one by one:

In Aryan’s AP: Common difference, d₁ = (- 2) – (- 5) = – 2 + 5 = 3

In Roshan’s:  AP: Common difference, d₂ = (184) – (187) = 184 – 187 = – 3

∴ Sum of the common differences = d₁ + d₂ = 3 + ( – 3) = 3 – 3 = 0

Therefore, the sum of the common differences of the two APs is 0.

ii. Value of the 34th term of Roshan’s AP:

In Roshan’s AP, we have first term, a = 187 and common difference, d = – 3

We know that nth term of an AP is given by, T_n = a + ( n – 1) d

Let’s calculate this value for n = 34:

∴ T₃₄ = 187 + (34 – 1) (- 3)

= 187 + 33 x (- 3)

= 187 – 99 = 88

Therefore, the value of the 34th term of Roshan’s AP is 88.

iii. Sum of first 10 terms of Aryan’s AP:

In Aryan’s AP, we have first term, a = – 5 and common difference, d = 3

sum of n terms, S_n = (n / 2) [2 a + (n + 1) d]

Let’s calculate this value for n = 10:

∴ S₁₀ = (10 / 2) [2 x (- 5) + (10 – 1) (3) ]

= 5 [ (- 10) + 9 (3)]

= 5 (- 10 + 27)

= 5 x 17 = 85

Therefore, the sum of first 10 terms of Aryan’s AP is 85.

iv. Value of the term with same value:

Let’s consider that nth term has same value in both APs.

We will write equations of nth term in both APs one by one and compare to get value of n.

We know that nth term of an AP is given by, T_n = a + ( n – 1) d

Let’s start from nth term of Aryan’s AP: 

T_nA = – 5 + (n – 1) (3)

= – 5 + 3 n – 3

= 3 n – 8 ………… (i)

Next, nth term of Roshan’s AP: 

T_nR = 187 + (n – 1) (- 3)

= 187 – 3 n + 3

= 190 – 3 n ………… (ii)

By comparing both equations (i) and (ii), we get:

3 n – 8 = 190 – 3 n

3 n + 3 n = 190 + 8

6 n = 198

n = 198 / 6 = 33

Therefore, at 33rd term, both APs will have same value:

Check: For Aryan’s AP, value of 33rd term = – 5 + (33 – 1) (3) = – 5 + 96 = 91
For Roshan’s AP, value of 33rd term = 187 + (33 – 1) (- 3) = 187 – 96 = 91
Since both values are equal, hence our answer is correct.

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