Question

Find two numbers such that the mean proportional between them is $12$ and third proportional to them is $96$.

Answer 1

Let the required no. be 'a' and 'b'

$\therefore$ is the mean proportional between a and b

$\therefore a:12=12:b$

$ab=144$.....(1)

96 is third proportional to a and b

$a:b=b:96$

$b^2=96a$....(2)

from 1 & 2, $ab=144$

$\Rightarrow a=\frac{144}{b}$

from 2, $b^2=96\times \frac{144}{b}$

$b^3=96\times 144$

$\Rightarrow b=24$

$\therefore a=\frac{144}{24}=6$

$\therefore a=6,b=24$