Question

The product of three numbers in G.P. is 125 and the sum of their products taken in pairs is $87\frac{1}{2}$. Find them.

Answer 1

Let the three numbers in G.P. be $\frac{a}{r},a,ar$

then product of these numbers $\left(\frac{a}{r}\right)\left(a\right)\left(ar\right)$

$\Rightarrow a^3=125=5^3$

$\Rightarrow a=5$

Also, sum of these products in pair

$\left(\frac{a}{r}\right)\left(a\right)+\left(a\right)\left(ar\right)+\left(\frac{a}{r}\right)\left(ar\right)=87\frac{1}{2}=\frac{175}{2}$

$\frac{a^2}{r}+a^{2}r+a^2=a^2(\frac{1}{r}+r+1)$

$\Rightarrow (5{)}^2\left(\frac{1+r^2+r}{r}\right)=\frac{175}{2}$

$1+r^2+r=\left(\frac{175}{2\times 25}\right)$

$2(1+r^2+r)=7r$

$\Rightarrow 2r^2+2r+2+7r$

$\Rightarrow 2r^2-5r+2=0$

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

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

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

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

Hence the G.P. for a = 2 and

$r=\frac{1}{2}is10,5,\frac{5}{2}$

And the G.P. for a = 5 and

r = 2 is $\frac{5}{2},5,10$