Question

is S1 S2 S3 are the sums of first n, 2n,3n terms of an AP prove that S3 = 3(s2 -S1)

Answer 1

We know that the sum of $n$ terms in an A.P. is 

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

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

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

Now,

R.H.S =

$3(S_{2} - S_{1})$

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

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

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

$= S_{3}$

Therefore the equality is proved.

Answer 2

plz refer to this attachment