Q) In an A.P., the first term is 4 and the last term is 31. If sum of all the terms is 175, find the number of terms and the common difference.

(Q 23A- 30/3/3 – CBSE 2026 Question Paper)

Ans:

Given that, first term, a = 4

Step 1: By given 1^st condition is given as “last term is 31”

∴ Last term = 31

Let’s consider the last term is n^th term

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

∴ Last term, 31 = (4) + (n – 1) d

∴ (n – 1) d = 31 – 4

∴ (n – 1) d = 27 ……….. (i)

Step 2: By given 1^st condition is given as “If sum of all the terms is 175”

∴ S_n = 175

We know that in an AP, Sum of n terms is given by:

S_n = [2a + (n – 1) d]

∴ 175 = [2 (4) + (n – 1) d] (∵ given a = 4)

∴ 175 x 2 = n (8 + (n – 1) d) ………… (ii)

Step 3: By substituting value of (n-1) d from equation (i), we get:

∴ 350 = n (8 + 27)

∴ 350 = 35 n

∴ n = = 10

Step 4: By substituting value of n in equation (i), we get:

∵ (n – 1) d = 27

∴ (10 – 1) d = 27

∴ 9 d = 27

∴ d = = 3

Therefore, number of terms are 10 and the common difference is 3.

Check: If a = 4, d = 3 and n = 10
Then last term or T₁₀ = a + (n – 1) d = 4 + (10-1) 3 = 4 + 9 x 3 = 4 + 27 = 31 ….. 1st condition macthed.
Sum of all terms or S₁₀ = [2a + (n – 1) d] = [2 (4) + (10 – 1) (3)] = 175 ….. 2nd condition matched
Since both the given conditions are matched, our answer is correct.

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