Question

the sum of the terms of AP whose 1st term ,last term and common difference are 3,101,and 7 respectively

Answer 1

First term (a) = 3

Last term (L) = 101

Common difference (d) = 7

So, required AP is

3, 10, 17, 24, 31, 38, 45, 52, 59, 66, 73, 80, 87, 94, 101

Here, number of terms (n) = 15

Sum of 15 terms of an AP is

S(n) = n/2(a + L)

S(15) = 15/2( 3 + 101 )

S(15) = 780 ______________ (Ans.)

Hope it helps.

^_°