The sum of first 15 terms of an A.P. is 105 and the sum of next 15 terms is 780. Find the A. P.

For an AP
with first term a and the common difference d, the sum of the first n terms is given as (2a +
(n-1)d)(n/2).

Now for the AP given the sum of the first 15 terms is
105

=> (2a + 14d)(15/2) = 105...(1)

The sum of the next
15 terms is the sum of the first 15 terms subtracted from the sum of the first 30
terms.

=> (2a + 29d)(30/2) - 105 = 780

=> (2a + 29d)
= 885/15

=> (2a + 29d) = 59

(1)

=> (2a + 14d)(15/2) = 105

=> 2a + 14d = 14

2a + 29d - 2a - 14d = 59 - 14

=> 15d = 45

=> d = 3

2a + 14d = 14

=> 2a = 14 -
42

=> 2a = -28

=> a = -14

Therefore the AP has the first term as -14 and the common difference is
3.