Question

foreign arithmetic progression if S M is equals to sn and then s m + n is equals to​

Answer 1

Answer:

$S_{m+n}=0$

Step-by-step explanation:

Formula used:

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

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

$Given:\\\\S_m=S_n\\\\\frac{m}{2}[2a+(m-1)d]=\frac{n}{2}[2a+(n-1)d]\\\\m[2a+(m-1)d]=n[2a+(n-1)d]\\\\2am+m^2d-md=2an+n^2d-nd$

Rearranging terms we get

$2am-2an+m^2d-n^2d-md+nd=0\\\\2a(m-n)+(m^2-n^2)d-(m-n)d=0\\\\2a(m-n)+(m+n)(m-n)d-(m-n)d=0$

Divide both sides by m-n

$2a+(m+n)d-d=0\\\\2a+((m+n)-1)d=0...........(1)$

$Now,\\\\S_{m+n}=\frac{m+n}{2}[2a+((m+n)-1)d]\\\\S_{m+n}=\frac{m+n}{2}[0]\:\:\:(using\:(1))\\\\S_{m+n}=0$