Question

the sum of n yerma , 2n terms and 3n terms of an AP are S1, S2 and S3 respectively. prove that S3=3(S2-S1).

Answer 1

Let a=first term

d=common difference.

S1=n/2*[2a+(n-1)d]...................(1)

S2=2n/2*[2a+(2n-1)d]....................(2)

S3=3n/2*[2a+(3n-1)d].......................(3)

S2-S1=2n/2*[2a+(2n-1)d] - n/2*[2a+(n-1)d].........{from(1) and (2)}

=n*[2a+(2n-1)d] - n/2*[2a+(n-1)d]

Taking n as common.

n[2a+(2n-1)d-1/2{2a+(n-1)d]

=n/2[4a+2d(2n-1)-2a-(n-1)d]

=n/2[2a+4nd-2d-dn+d]

=n/2[2a+3nd-d]

=n/2[2a+d[3n-1]

Now S3=3n/2*[2a+(3n-1)d]...............(see(3))

S2-S1==n/2[2a+d[3n-1]

3(S2-S1)=3n/2*[2a+(3n-1)d]

=S3

Hence proved.

Hope it helps you.

Answer 2

plz refer to this attachment