Question

Prove that an=Sn-S(n-1) in AP

Answer 1

Answer:

First u prove it

Step-by-step explanation:

Then I will.prove

Answer 2

use relation of $n^{th}$  term and relation of sum of n terms

Step-by-step explanation:

$a_{n}= a+(n-1)d$ the $n^{th}$ term relation

$S_{n}=\frac{n}{2}[2a+(n-1)d]$ is the relation for sum of n terms

$S_{n-1}=\frac{n-1}{2}[2a+(n-1-1)d]$ is the relation for sum of $(n-1)$ terms

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

After open some brackets,

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

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

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

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

After cancellation same terms having apposite signs,

                             $=a+dn-d$

or,     $S_{n}-S_{(n-1)}=a+(n-1)d=a$