Question

If $A$ and $G$ are the Arithmetic mean and Geometric mean respectively of two numbers, then prove that the numbers are $A\pm \sqrt{(A+G)(A-G)}$.

Answer 1

Let the two numbers be $a-d$ and $a+d$

Arithmetic mean of $a-d$ and $a+d$ is: $A=\frac{a-d+a+d}{2}=a$

Geometric Mean of $a-d$ and $a+d$ is: $G=\sqrt{(a-d)(a+d)}=\sqrt{a^2-d^2}$

Now,

$A=a$ ... (i)

$G^2=a^2-d^2$ ... (ii)

Subtracting (ii) from the square of (i), we get

$A^2-G^2=a^2-(a^2-d^2)$

$\Rightarrow (A-G)(A+G)=(d{)}^2$

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

and $a=A$

Therefore,

Numbers are $A\pm \sqrt{(A-G)(A+G)}$