Answer:

Given that 15

th

term is 15 more than the 17 th term

a+14d=15+a+11d

14d−11d=15

3d=15

d=5

Substitute the value of d in t

8term we get

a+7(5)=37

a+35=37

a=2

a=2, d=5

Now, AP terms will be

First term a=2

2nd

term a+d=7

3 rd

term −a+2d=12

4 th

term =a+3d=17 and so on

AP =2,7,12,17….

Sum of 1st

20 term in AP=s

n =n/2(2a+(n−1)d)

=20/2[2×2+(20−1)×5]

=10[4+19×5]=10[4+95]=10×99=990