# 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.

*print*Print*list*Cite

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.**

The sum of the n terms of an AP with starting term a1 and common difference is given bY:Sn = {2a+(n-1)d}n/2.

Therefore the sum of the first 15 terms is S15 = {2a+15-1)d}/15/2 is given to be 105.

So s15 = {2a+14d)*15/2 = 105

=> 2a+ 14d)15 = 105 *2 = 210.

2a+14d =210/15 = 14.

2a+14d = 14.....(1)

Since the sum of the next 15 terms = 780,

S30 - s15 = 730.

=> {2a+29d)30/2 = 105+780 = 105+780 =885.

=> 2a+29d = 885*2/30 = 59.

=> 2a+29d = 59.....(2).

(2)-1): 29d-14d = 59-14 = 45.

15d = 45.

d= 45/15 = 3. Substituting this value of d = 5/3, in 2a+14d = 14, we get, 2a = 14-(14)*3 = -28. So a = -14.

So a = -14

Therefore the required A.P. has the starting term - 14 and the common difference 3.