Q) In an A.P., 15 ^th term exceeds the 8 ^th term by 21. If sum of first 10 terms is 55, then form the A.P. 

(Q 29 A – 30/5/2 – CBSE 2026 Question Paper)

Ans:

Let’s consider the first term is a and Common difference is d

Step 1: By 1 ^st given condition, “15 ^th term exceeds the 8 ^th term by 21″

T₁₅ – T₈ = 21 ………. (i)

Step 2: We know that the n ^th term of an AP is given by: T_n = a + (n – 1) d

∴ 15 ^th term of the AP, T₁₅ = a + (15 – 1)d

∴ T₁₅ = a + 14 d

Similarly, 8 ^th term of this AP, T₈ = a + (8 – 1) d

∴ T₈ = a + 7 d

∴ T₁₅ – T₈ = (a + 14 d) – (a + 7 d)

= a + 14 d – a – 7 d = 7 d

Step 3: Substituting values of (T₁₅ – T₈) in equation (i), we get:

∴ 7 d = 21

∴ d = = 3

Step 4: By 2 ^nd given condition, “Sum of first 10 terms is 55”

∴ S₁₀ = 55 ………..(ii)

We know that sum of first n terms of an AP is given by:

S_n = [2 a + (n – 1) d]

∴ S₁₀ = [2 a + (10 – 1) (3)]        (∵ d = 3 calculated in step 3)

∴ S₁₀ = 5 (2 a + 27)

∴ S₁₀ = 10 a + 135 …. (iii)

Step 5: By comparing equations (ii) and (iii), we get:

10 a + 135 = 55

∴ 10 a = 55 – 135 = – 80

∴ a = = – 8

Step 6: Now, for the AP, we have:

first term a = – 8

common difference, d = 3

Hence, nth term, T_n = a + (n – 1) d = (- 8) + (n – 1) (3)

∴ T_n =- 8 + 3 n – 3 =  3 n – 11

∴ value of 2 ^nd term = 3 (2) – 11 = 6 – 11 = – 5

and value of 3 ^th term = 3 (3) – 11 = 9 – 11 = – 2

and value of 4 ^th term = 3 (4) – 11 = 12 – 11 = 1

Therefore, here AP is – 8, – 5, – 2, 1…..

Check: at a = – 8 an d = 3, let’s check the given conditions one by one:
1. T₁₅ – T₈ = 7 d = 7 (3) = 21. It meets 1 ^st condition.
2. S₁₀ = 10 a + 135 = 10 (-8) – 135 = – 80 + 135 = 55. It meets 2 ^nd condition.
Since both the given condition are met, our solution is correct.

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