Question

If S₁, S₂, S₃ be respectively the sums of n, 2n, 3n terms of a G.P., then prove that $S_1^2+S_2^2$ = S₁ (S₂ + S₃).

Answer 1

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

$$\begin{gathered} \mathrm{Sum}\mathrm{of}\mathit{}n\mathrm{terms},S_1=a\left(\frac{r^n-1}{r-1}\right)...\left(1\right) \\ \mathrm{Sum}\mathrm{of}\mathit{}2n\mathrm{terms},S_2=a\left(\frac{r^{2n}-1}{r-1}\right) \\ \Rightarrow S_2=a\left[\frac{{\left(r^n\right)}^2-1^2}{r-1}\right] \\ \Rightarrow S_2=a\left[\frac{\left(r^n-1\right)\left(r^n+1\right)}{r-1}\right] \\ \Rightarrow S_2=S_1\left(r^n+1\right)....\left(2\right) \\ \mathrm{And},s\mathrm{um}\mathrm{of}\mathit{}3n\mathrm{terms},S_3=a\left(\frac{r^{3n}-1}{r-1}\right) \\ \Rightarrow S_3=a\left[\frac{{\left(r^n\right)}^3-1^3}{r-1}\right] \\ \Rightarrow S_3=a\left[\frac{\left(r^n-1\right)\left(r^{2n}+r^n+1\right)}{r-1}\right] \\ \Rightarrow S_3=S_1\left(r^{2n}+r^n+1\right)...\left(3\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\mathrm{LHS}={\left(S_1\right)}^2+{\left(S_2\right)}^2 \\ ={\left(S_1\right)}^2+{\left[S_1\left(r^n+1\right)\right]}^2\left[\mathrm{Using}\left(2\right)\right] \\ ={\left(S_1\right)}^2\left[1+{\left(r^n+1\right)}^2\right] \\ ={\left(S_1\right)}^2\left[1+r^{2n}+2r^n+1\right] \\ ={\left(S_1\right)}^2\left[r^{2n}+r^n+1+r^n+1\right] \\ =\left(S_1\right)\left[S_1\left(r^{2n}+r^n+1\right)+S_1\left(r^n+1\right)\right] \\ =\left(S_1\right)\left[S_2+S_3\right]\left[\mathrm{Using}\left(2\right)\mathrm{and}\left(3\right)\right] \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$