Question

Find two numbers whose arithmetic mean is 34 and the geometric mean is 16.

Answer 1

Let the two numbers be $a$ and $b$ such that $a>b$

Then,

$\frac{a+b}{2}=34$ and $\sqrt{ab}=16$

$a+b=68$.......(1) and $ab=256$

$\therefore (a-b{)}^2=(a+b{)}^2-4ab$

$(a-b{)}^2=(68{)}^2-4\times 256=3600$

$a-b=60$.......(2)

solving eqn (1) & (2), we get,

$a=64$ and $b=4$

Hence, required numbers are $64$ and $4$.