Question

If the product of 3 numbers in GP is 125 & sum of their products taken in pairs is $87\frac{1}{2}$ then find the numbers.

Answer 1

Solution:

Let the three numbers are $\frac{a}{r},a,ar$

produced $=125$

$\Rightarrow \frac{a}{r}\times a\times ar=125$

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

Now also sum of their products taken in pairs is $87\frac{1}{2}$

$\therefore [\frac{a}{r}\times a+a.ar+ar\times \frac{a}{r}]=87\frac{1}{2}$

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

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

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

$\Rightarrow 1+r^2+r=\frac{7}{2}r$

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

$\Rightarrow r=\frac{5\pm \sqrt{25-16}}{4}=\frac{5\pm 3}{4}=2,\frac{1}{2}$

For $r=2,a=5$

the numbers are $\frac{a}{r},a,ar\Rightarrow \frac{5}{2},5,10$

For $r=\frac{1}{2}$

the numbers are $\frac{a}{r},a,ar\Rightarrow 10,5,\frac{5}{2}$

$\therefore$ Number are $\frac{5}{2},5,10$