Question

If sn denotes the sum of n terms of an ap then the value of s2n-sn is equal to

Answer 1

Answer:

$S_{2n}-S_n=\frac{1}{3}S_{3n}$

Step-by-step explanation:

Formula used:

The sum of n terms of an A.P a, a+d, a+2d,.... is

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

consider,

$S_{2n}-S_n$

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

$=\frac{1}{2}[4an+(4n^2-2n)d]-\frac{1}{2}[2an+(n^2-n)d]$

$=\frac{[4an+4n^2d-2nd]-[2an+n^2d-nd]}{2}$

$=\frac{4an+4n^2d-2nd-2an-n^2d+nd}{2}$

$=\frac{2an+3n^2d-nd}{2}$

$=\frac{n}{2}[2a+3nd-d]$

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

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

$=(\frac{1}{3})S_{3n}$

$\implies\:\bold{S_{2n}-S_n=\frac{1}{3}S_{3n}}$