Question

if S1,S2 and S3 be respectively the sum of n,2n and 3n terms of gp prove that S1 (S3-S2)=(S2-S1)^2

Answer 1

hence proved.
it was a very interesting question

Answer 2

From the given we haved proved that $S_1(S_3-S_2)=(S_2-S_1)^2$

Step-by-step explanation:

Given that $S_1,S_2,S_3$ be respectively the sum of n,2n and 3n terms of GP

To prove that $S_1(S_3-S_2)=(S_2-S_1)^2$ :

• First take LHS $S_1(S_3-S_2)$

Substitute the values we get

$=n(3n-2n)$

$=n(n)$

$=n^2$

Therefore $S_1(S_3-S_2)=n^2=LHS$

• Then RHS $(S_2-S_1)^2$

Substitute the values we get

$=(2n-n)^2$

$=n^2$

Therefore  $(S_2-S_1)^2=n^2=RHS$

Implies that LHS=RHS

Therefore $S_1(S_3-S_2)=(S_2-S_1)^2$

Hence proved

From the given we haved proved that $S_1(S_3-S_2)=(S_2-S_1)^2$