Question

If the sum of first n ,2n and 3n terms of AP be s1, s2 and s3 respectively,then prove that s3=3(s2-s1).

Answer 1

I think this will help I didn't clarify it

Answer 2

Answer:

$3(S_2-S_1)=S_3$

Step-by-step explanation:

It is given that the sum of first n ,2n and 3n terms of AP be s1, s2 and s3.

To prove: $S_3=3(S_2-S_1)$

The sum of first n terms is defined as

$S_1=S_n=\frac{n}{2}[2a+(n-1)d]$

$S_2=S_{2n}=\frac{2n}{2}[2a+(2n-1)d]$

$S_3=S_{3n}=\frac{3n}{2}[2a+(3n-1)d]$

Taking right hand side of the given equation.

$3(S_2-S_1)=3(\frac{2n}{2}[2a+(2n-1)d]-\frac{n}{2}[2a+(n-1)d])$

$3(S_2-S_1)=\frac{3n}{2}(2[2a+2dn-d]-[2a+dn-d])$

$3(S_2-S_1)=\frac{3n}{2}(4a+4dn-2d-2a-dn+d)$

$3(S_2-S_1)=\frac{3n}{2}(2a+3dn-d)$

$3(S_2-S_1)=\frac{3n}{2}(2a+(3n-1)d)$

$3(S_2-S_1)=S_{3n}$

$3(S_2-S_1)=S_3$

L.H.S=R.H.S.

Hence prove.