Question

Can there be two distinct numbers such that their arithmetic mean (A.M.) $=$ geometric mean (G.M).

Answer 1

No, there can not be two distinct numbers satisfying the given conditions.

Proof by contradiction:-

If possible let there are two numbers $a,b>0$ with $a\ne b$ such that A.M. and G.M. of them are equal.

Then $\frac{a+b}{2}=\sqrt{ab}$

or,$(a+b{)}^2=4ab$ [ Squaring both sides we get]

or,$(a-b{)}^2=0$

or, $a=b$.

Which is a contradiction to the fact that $a\ne b$.