Question

the sum of first n,2n and 3n terms of an A.p are s1,s2 and s3 respectively. prove that s3 = 3 (s2-s1).​

Answer 1

Answer:

Given:

$\sf{Sum \ of \ n \ terms=S_{1},}$

$\sf{Sum \ of \ 2n \ terms=S_{2},}$

$\sf{Sum \ of \ 3n \ terms=S_{3}}$

To prove:

$\sf{S_{3}=3(S_{2}-S_{1})}$

Proof:

$\sf{L.H.S.=S_{3}}$

$\sf{=S_{3n}}$

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

____________________

$\sf{R.H.S.=3(S_{2}-S_{1})}$

$\sf{=3(S_{2n}-S_{n})}$

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

$\sf{=3\{(n[2a+2nd-d])-(n[a+\dfrac{nd}{2}-\dfrac{d}{2}])\}}$

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

$\sf{=3\{n[a+\dfrac{4nd-nd}{2}-\dfrac{2d+d}{2}]\}}$

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

$\sf{Taking \ \dfrac{1}{2} \ as \ common}$

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

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

$\sf{=L.H.S.}$

$\sf{Hence \ proved,}$

$\purple{\tt{S_{3}=3(S_{2}-S_{1})}}$

Answer 2

plz refer to this attachment