Question

If S denotes the sum of an infinite G.P. S₁ denotes the sum of the squares of its terms, then prove that the first term and common ratio are respectively $\frac{2S{S}_1}{S^2+S_1}\mathrm{and}\frac{S^2-S_1}{S^2+S_1}$.

Answer 1

$$\begin{gathered} \mathrm{S}=\frac{a}{\left(1-r\right)}.......\left(\mathrm{i}\right) \\ \mathrm{And},{\mathrm{S}}_1=\frac{a^2}{\left(1-r^2\right)} \\ \Rightarrow {\mathrm{S}}_1=\frac{a^2}{\left(1-r\right)\left(1+r\right)}.......\left(\mathrm{ii}\right) \\ \mathrm{Now},\mathrm{putting}\mathrm{the}\mathrm{value}\mathrm{of}a\mathrm{in}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right): \\ {\mathrm{S}}_1=\frac{{\mathrm{S}}^2{\left(1-r\right)}^2}{\left(1-r\right)\left(1+r\right)} \\ \Rightarrow {\mathrm{S}}_1=\frac{{\mathrm{S}}^2\left(1-r\right)}{\left(1+r\right)} \\ \Rightarrow {\mathrm{S}}_1\left(1+r\right)={\mathrm{S}}^2\left(1-r\right) \\ \Rightarrow r\left({\mathrm{S}}_1+{\mathrm{S}}^2\right)={\mathrm{S}}^2-{\mathrm{S}}_1 \\ \Rightarrow r=\frac{\left({\mathrm{S}}^2-{\mathrm{S}}_1\right)}{\left({\mathrm{S}}_1+{\mathrm{S}}^2\right)} \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathit{}r\mathit{}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right): \\ \Rightarrow a=\mathrm{S}\left(1-r\right) \\ \Rightarrow a=\mathrm{S}\left(1-\frac{\left({\mathrm{S}}^2-{\mathrm{S}}_1\right)}{\left({\mathrm{S}}_1+{\mathrm{S}}^2\right)}\right) \\ \Rightarrow a=\mathrm{S}\left(\frac{\left({\mathrm{S}}_1+{\mathrm{S}}^2\right)-\left({\mathrm{S}}^2-{\mathrm{S}}_1\right)}{\left({\mathrm{S}}_1+{\mathrm{S}}^2\right)}\right) \\ \Rightarrow a=\frac{2{\mathrm{SS}}_1}{\left({\mathrm{S}}_1+{\mathrm{S}}^2\right)} \end{gathered}$$