Question

If $S_1,S_2,and{S}_3$ are respectively the sum of n,2n and 3n terms of a G.P, then $S_1(S_3-S_2)=(S_2-S_1{)}^2$.

• A. True
• B. False

Answer 1

The correct option is A True

Let a be the first term and r be the common ratio of the G.P.

$S_1$ = Sum of n terms of G.P. = $\frac{a(r^n-1)}{r-1}$

$S_2$ = Sum of 2n terms of G.P. = $\frac{a(r^{2n}-1)}{r-1}$

$S_3$ = Sum of 3n terms of G.P. = $\frac{a(r^{3n}-1)}{r-1}$

L.H.S. = $S_1(S_3-S_2)=\frac{a(r^n-1)}{r-1}[\frac{a(r^{3n}-1)}{r-1}-\frac{a(r^{2n}-1)}{r-1}]$

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

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

$=\frac{a^2{r}^{2n}(r^n-1{)}^2}{(r-1{)}^2}$

R.H.S. = $(S_2-S_1{)}^2=[\frac{a(r^{2n}-1)}{r-1}-\frac{a(r^n-1)}{r-1}{]}^2$

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

$=\frac{a^2}{(r-1{)}^2}[r^{2n}-r^n{]}^2$

$=\frac{a^2{r}^{2n}}{(r-1{)}^2}(r^n-1{)}^2$

Therefore, LHS = RHS