Question

If the arithmetic and geometric mean of two positive real numbers $a$ and $b(a>b)$ are $13$ and $12$ respectively, then the value of $a-b$ is

• A. $10$
• B. $12$
• C. $13$
• D. $8$

Answer 1

The correct option is A $10$

Given, $AM=13$ and $GM=12$
$\Rightarrow \frac{a+b}{2}=13$ and $\sqrt{ab}=12$
$\Rightarrow a+b=26$ and $\sqrt{ab}=12...\left(1\right)$
Squaring both sides, we get
$(a+b{)}^2=676$ and $ab=144$
$\Rightarrow (a-b{)}^2+4ab=676$
$\Rightarrow (a-b{)}^2=676-4\times 144=100$
$\Rightarrow a-b=10(\because a>b)...\left(2\right)$

From $\left(1\right)$ and $\left(2\right),$ we get
$a=18,b=8$
$\therefore a-b=10$