# 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.

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

Sn = [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

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