Solution:
It is given that A.M = 25, G.M = 20.
We have, G.M = √ab
and A.M = (a+b)/2
So, we can write:
√ab = 20 ……. (1)
(a+b)/2 = 25……. (2)
Also, a + b = 50
Or, a = 50 – b
Substituting the value of ‘a’ in equation (1), we get,
$ \surd \left[ \left( 50-b \right)b \right]\text{ }=\text{ }20 $
$ 50b-{{b}^{2}}~=\text{ }400 $
$ {{b}^{2}}-50b\text{ }+\text{ }400\text{ }=\text{ }0 $
$ {{b}^{2}}-40b-10b\text{ }+\text{ }400\text{ }=\text{ }0 $
$ b\left( b-40 \right)-10\left( b-40 \right)\text{ }=\text{ }0 $
$ b\text{ }=\text{ }40\text{ }or\text{ }b\text{ }=\text{ }10 $
When b = 40 then a = 10
When b = 10 then a = 40
Therefore, the numbers are 10 and 40.