Question

If the sum of an infinite decreasing G.P. is 3 and the sum of the squares of its term is $\frac{9}{2}$, then write its first term and common difference.

Answer 1

Let us take a G.P. whose first term is a and common difference is r.
$$\begin{gathered} \therefore S_{\infty }=\frac{a}{1-r} \\ \Rightarrow \frac{a}{1-r}=3.......\left(\mathrm{i}\right) \\ \mathrm{And},\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{terms}\mathrm{of}\mathrm{the}\mathrm{G}.\mathrm{P}.a^2,{\left(ar\right)}^2,{\left(a{r}^2\right)}^2,...\infty : \\ S{\text{'}}_{\infty }=\frac{a^2}{1-r^2} \\ \Rightarrow \frac{a^2}{1-r^2}=\frac{9}{2}.......\left(\mathrm{ii}\right) \\ \Rightarrow 2a^2=9\left(1-r^2\right) \\ \Rightarrow 2{\left[3\left(1-r\right)\right]}^2=9-9r^2\left[\mathrm{From}\left(\mathrm{i}\right)\right] \\ \Rightarrow 18\left(1+r^2-2r\right)=9-9r^2 \\ \Rightarrow 18-9+18r^2+9r^2-36r=0 \\ \Rightarrow 27r^2-36r+9=0 \\ \Rightarrow 3\left(9r^2-12r+3\right)=0 \\ \Rightarrow 9r^2-12r+3=0 \\ \Rightarrow 9r^2-9r-3r+3=0 \\ \Rightarrow 9r\left(r-1\right)-3\left(r-1\right)=0 \\ \Rightarrow \left(9r-3\right)\left(r-1\right)=0 \\ \Rightarrow r=\frac{1}{3}andr=1. \\ \mathrm{But},r=1\mathrm{is}\mathrm{not}\mathrm{possible}. \\ \therefore r=\frac{1}{3} \\ \mathrm{Now},\mathrm{putting}r=\frac{1}{3}\mathrm{in}\frac{\mathrm{a}}{1-\mathrm{r}}=3: \\ \mathrm{a}=3\left(1-\frac{1}{3}\right) \\ \Rightarrow \mathrm{a}=3\times \frac{2}{3}=2 \end{gathered}$$