The product of three numbers in G.P.

Solution:

Let the required numbers be $\frac{a}{r}, a$ and $a r$.

Product of the G.P. = 125

$\Rightarrow a^{3}=125$

$\Rightarrow a=5$

Sum of the products in pairs $=87 \frac{1}{2}=\frac{175}{2}$

$\Rightarrow \frac{a}{r} \times a+a \times a r+a r \times \frac{a}{r}=\frac{175}{2}$

$\Rightarrow \frac{a^{2}}{r}+a^{2} r+a^{2}=\frac{175}{2}$

Subs tituting the value of $a$

$\Rightarrow \frac{25}{r}+25 r+25=\frac{175}{2}$

$\Rightarrow 50 r^{2}+50 r+50=175 r$

$\Rightarrow 50 r^{2}-125 r+50=0$

$\Rightarrow 25\left(2 r^{2}-5 r+2\right)=0$

$\Rightarrow 2 r^{2}-4 r-r+2=0$

$\Rightarrow 2 r(r-2)-1(r-2)=0$

$\Rightarrow(2 r-1)(r-2)=0$

$\therefore r=\frac{1}{2}, 2$

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 .