Question

the first term of an arthimetic sequances is 1 and the sum of first four terms is 100. find the first four terms
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Answer 1

$\huge\bold\star\purple{Answer}$

Let a, a+d, a+2d, a+3d be the 4 terms of Ap a=1, a+(a+d) +(a+2d) +(a+3d) =100

1+(1+d)+(1+2d)+(1+3d)=100

1+1+d+1+2d+1+3d=100

4+6d=100

6d=100-4

6d=96

d=96÷6

d=16

a=1,d=16

Ap=a,a+d,a+2d,a+3d

Ap=1,1+16,1+2(16),1+3(16)

Ap=1,17,1+32,1+48

Ap=1,17,33,49.......☺