Question

The sum of an infinite G.P. is 4 and the sum of the cubes of its terms is 92. The common ratio of the original G.P. is
(a) 1/2
(b) 2/3
(c) 1/3
(d) −1/2

Answer 1

(a) 1/2

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{G}.\mathrm{P}.\mathrm{be}a,ar,a{r}^2,a{r}^3,...,\infty . \\ {S}_{\infty }=4 \\ \Rightarrow \frac{a}{1-r}=4\left(\mathrm{i}\right) \\ \mathrm{Also},\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{cubes},S_1=92 \\ \Rightarrow \frac{a^3}{\left(1-r^3\right)}=92\left(\mathrm{ii}\right) \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}a\mathrm{from}\left(\mathrm{i}\right)\mathrm{to}\left(\mathrm{ii}\right): \\ \Rightarrow \frac{{\left(4(1-r)\right)}^3}{\left(1-r^3\right)}=92 \\ \Rightarrow \frac{64(1-r{)}^3}{\left(1-r^3\right)}=92 \\ \Rightarrow \frac{{\left(1-r\right)}^3}{\left(1-r\right)\left(1+r+r^2\right)}=\frac{92}{64} \\ \Rightarrow \frac{{\left(1-r\right)}^2}{\left(1+r+r^2\right)}=\frac{23}{16} \\ \Rightarrow 16\left(1-2r+r^2\right)=23\left(1+r+r^2\right) \\ \Rightarrow 7r^2+55r+7=0 \\ \mathrm{Using}\mathrm{the}\mathrm{quadratic}\mathrm{formula}: \\ \Rightarrow r=\frac{-55+\sqrt{{55}^2-4\times 7\times 7}}{2\times 7} \\ \Rightarrow r=\frac{-55+\sqrt{{55}^2-{14}^2}}{14} \\ \Rightarrow r=\frac{-55+\sqrt{2829}}{14} \end{gathered}$$

Disclaimer: None of the given options are correct. This solution has been created according to the question given in the book.