Question

Mark the correct alternative in the following question:

Let S be the sum, P be the product and R be the sum of the reciprocals of 3 terms of a G.P. Then p²R³ : S³ is equal to

(a) 1 : 1 (b) (Common ratio)ⁿ : 1 (c) (First term)² : (Common ratio)² (d) None of these

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{three}\mathrm{terms}\mathrm{of}\mathrm{the}\mathrm{G}.\mathrm{P}.\mathrm{be}\frac{a}{r},a,ar.\mathrm{Then} \\ S=\frac{a}{r}+a+ar \\ =a\left(\frac{1}{r}+1+r\right) \\ =a\left(\frac{1+r+r^2}{r}\right) \\ =\frac{a\left(r^2+r+1\right)}{r} \\ \mathrm{Also}, \\ P=\frac{a}{r}\times a\times ar=a^3 \\ \mathrm{And}, \\ R=\frac{r}{a}+\frac{1}{a}+\frac{1}{ar} \\ =\frac{1}{a}\left(r+1+\frac{1}{r}\right) \\ =\frac{1}{a}\left(\frac{r^2+r+1}{r}\right) \\ \mathrm{Now}, \\ \frac{P^2{R}^3}{S^3}=\frac{{\left(a^3\right)}^2\times {\left[{\displaystyle \frac{1}{a}}\left({\displaystyle \frac{r^2+r+1}{r}}\right)\right]}^3}{{\left[a\left({\displaystyle \frac{r^2+r+1}{r}}\right)\right]}^3} \\ =\frac{a^6\times {\displaystyle \frac{1}{a^3}}{{\displaystyle \left(\frac{r^2+r+1}{r}\right)}}^3}{a^3{\left({\displaystyle \frac{r^2+r+1}{r}}\right)}^3} \\ =\frac{1}{1} \\ \mathrm{So},\mathrm{the}\mathrm{ratio}\mathrm{is}1:1. \end{gathered}$$

Hence, the correct alternative is option (a).