Question

Let a1,a2,a3,....., an ...... be in A.P. If
a3+ a7+ a11 +a15 = 72, then the sum of its first 17 terms is equal to :

(a) 306
(b) 153
(c)612
(d) 204​

Answer 1

Given-

a3+a7+a11+a15= 72

a+2d+a+6d+a+10d+a+14d= 72

4a+32d= 72.

a+8d= 18. equation (1)

Let sum of first 17th term be s

a1+a2+a3+a4+........+a16+a17 = s

a+a+d+a+2d+a+3d+........+a+16d = s

17a+136d = s

Multiply equation (1) from 17 , we get

17a+136b = 306

so,sum of 17th terms of this AP is 306.