Question

The sum of the first 15 terms of an arithmetic progression is 105 and the sum of the next 15 terms is 780. Find the first three terms of the arithmetic progression.

Answer 1

Given,

Sum of first 15 terms = 105

Sum of next 15 terms = 780

Let us assume that there are 30 terms in AP

Therefore,sum of first 30 terms of AP = 105+780= 885

Now,

S30= n/2(2a+(n-1)d)

885=30/2{2a+(30-1)d}

2a+29d= (885*2)/30

2a+29d=59----------------1

Now,

S15=n/2{2a+(n-1)d}

105=15/2{2a+(15-1)d}

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

2a+14d=14 -------------------2

Now,

Subract eq 2from 1 , we get

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

2a+29d-2a-14d=45

15d=45

d=3

Put d=3in eq 2, we get

2=> 2a+14(3)=14

=>2a=14-42

=>a= -28/2

=>a=-14

Hence the first three terms of an AP is -14,-11,-8,............