The sum of the 4th and the 8th terms of an AP is 24 and the sum of its 6th and 10th terms is 44. Find the sum of its first 10 terms. (CBSE 2012)

Let a be the first term and d be the common difference.

Given: a₄ + a₈ = 24

and a₆ + a₁₀ = 44

To find: S₁₀

Now, Consider a₄ + a₈ = 24

⇒ a + 3d + a + 7d= 24

⇒ 2a + 10d = 24 ………….(1)

Consider a₆ + a₁₀ = 44

⇒ a + 5d + a + 9d = 44

⇒ 2a + 14d = 44 ………..(2)

Subtracting equation (1) from equation (2), we get,

4d = 20

⇒ d = 5

∴ Common difference = d = 5

Thus from equation (1), we get,

a = - 13

Now, Sum of first n terms of an AP is

S_n = [2a + (n - 1)d]

∴ Sum of first 10 terms is given by:

S₁₀ = [2(-13) + (10 - 1)(5)]

= 5 × [ - 26 + 45]

= 5 × 19

= 95

∴ S₁₀ = 95