Question

If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.

Answer 1

Let a and d be the first term and common difference, respectively of an AP.

∵ Sum of n terms of an AP,

S_n = `n/2 [2a + (n - 1)d]`  ...(i)

Now, S₆ = 36  ...[Given]

⇒ `6/2[2a + (6 - 1)d]` = 36

⇒ 2a + 5d = 12  ...(ii)

And S₁₆ = 256

⇒ `16/2[2a + (16 - 1)d]` = 256

⇒ 2a + 15d = 32  ...(iii)

On subtracting equation (ii) from equation (iii), we get

10d = 20

⇒ d = 2

From equation (ii),

2a + 5(2) = 12

⇒ 2a = 12 − 10 = 2

⇒ a = 1

∴ S₁₀ = `10/2 [2a + (10 - 1)d]`

= 5[2(1) + 9(2)] 

= 5(2 + 18)

= 5 × 20

= 100

Hence, the required sum of first 10 terms is 100.