Question

The product of three numbers in G.P. is 125 and the sum of their products taken in pairs is $87\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$. Find them.

Answer 1

Let the required numbers be $\frac{a}{r}\mathit{,}\mathit{}a\mathit{}\mathrm{and}\mathit{}ar.$
Product of the G.P. = 125
$$\begin{gathered} \Rightarrow a^3=125 \\ \Rightarrow \mathit{}a=5 \end{gathered}$$
Sum of the products in pairs = $87\frac{1}{2}=\frac{175}{2}$

$$\begin{gathered} \Rightarrow \frac{a}{r}\times a+a\times ar+ar\times \frac{a}{r}=\frac{175}{2} \\ \Rightarrow \frac{a^2}{r}+a^{2}r+a^2=\frac{175}{2} \\ \mathrm{Subs}\mathrm{tituting}\mathrm{the}\mathrm{value}\mathrm{of}a \\ \Rightarrow \frac{25}{r}+25r+25=\frac{175}{2} \\ \Rightarrow 50r^2+50r+50=175r \\ \Rightarrow 50r^2-125r+50=0 \\ \Rightarrow 25(2r^2-5r+2)=0 \\ \Rightarrow 2r^2-4r-r+2=0 \\ \Rightarrow 2r(r-2)-1(r-2)=0 \\ \Rightarrow (2r-1)(r-2)=0 \\ \therefore r=\frac{1}{2},2 \end{gathered}$$

Hence, the G.P. for a = 5 and r = $\frac{1}{2}$ is 10, 5 and $\frac{5}{2}$.
And, the G.P. for a = 5 and r = 2 is $\frac{5}{2}$, 5 and 10.