Question

Show that in an infinite G.P. with common ratio r (|r| < 1), each term bears a constant ratio to the sum of all terms that follow it.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{us}\mathrm{take}\mathrm{a}\mathrm{G}.\mathrm{P}.\mathrm{with}\mathrm{terms}{a}_1,a_2,a_3,a_4,...\infty \mathrm{and}\mathrm{common}\mathrm{ratio}r\left(\left|r\right|<1\right). \\ \mathrm{Also},\mathrm{let}\mathrm{us}\mathrm{take}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{the}\mathrm{terms}\mathrm{following}\mathrm{each}\mathrm{term}\mathrm{to}\mathrm{be}{\mathrm{S}}_1,{\mathrm{S}}_2,{\mathrm{S}}_3,{\mathrm{S}}_4,... \\ \mathrm{Now},{\mathrm{S}}_1=\frac{a_2}{\left(1-r\right)}=\frac{ar}{\left(1-r\right)}, \\ {S}_2=\frac{a_3}{\left(1-r\right)}=\frac{a{r}^2}{\left(1-r\right)}, \\ {S}_3=\frac{a_4}{\left(1-r\right)}=\frac{a{r}^3}{\left(1-r\right)}, \\ . \\ . \\ . \\ \Rightarrow \frac{a_1}{S_1}=\frac{a}{\frac{ar}{\left(1-r\right)}}=\frac{\left(1-r\right)}{r}, \\ \frac{a_2}{S_2}=\frac{ar}{\frac{a{r}^2}{\left(1-r\right)}}=\frac{\left(1-r\right)}{r}, \\ \frac{a_3}{S_3}=\frac{a{r}^2}{\frac{a{r}^3}{\left(1-r\right)}}=\frac{\left(1-r\right)}{r}, \\ . \\ . \\ . \\ \mathrm{It}\mathrm{is}\mathrm{clearly}\mathrm{seen}\mathrm{that}\mathrm{the}\mathrm{ratio}\mathrm{of}\mathrm{each}\mathrm{term}\mathrm{to}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{the}\mathrm{terms}\mathrm{following}\mathrm{it}\mathrm{is}\mathrm{constant}. \end{gathered}$$