If A and G be A.M. and G.M., respectively between two positive numbers,

Solution:

It is given that A and G are A.M. and G.M. between two positive numbers. Let these two positive numbers be a and b.

$\therefore \mathrm{AM}=\mathrm{A}=\frac{\mathrm{a}+\mathrm{b}}{2}$ $\ldots(1)$

$\mathrm{GM}=\mathrm{G}=\sqrt{\mathrm{ab}}$ ..(2)

From $(1)$ and $(2)$, we obtain

$a+b=2 A \ldots$ (3)

$a b=G^{2} \ldots(4)$

Substituting the value of $a$ and $b$ from $(3)$ and $(4)$ in the identity $(a-b)^{2}=(a+b)^{2}-4 a b$, we obtain

$(a-b)^{2}=4 A^{2}-4 G^{2}=4\left(A^{2}-G^{2}\right)$

$(a-b)^{2}=4(A+G)(A-G)$

$(a-b)=2 \sqrt{(A+G)(A-G)}$ ..(5)

From (3) and (5), we obtain

$2 \mathrm{a}=2 \mathrm{~A}+2 \sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}$

$\Rightarrow a=A+\sqrt{(A+G)(A-G)}$

Substituting the value of a in (3), we obtain

$\mathrm{b}=2 \mathrm{~A}-\mathrm{A}-\sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}=\mathrm{A}-\sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}$

Thus, the two numbers are $\mathrm{A} \pm \sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}$.