# Question e2cba

Sep 5, 2016

The reqd. nos. are $8 \mathmr{and} 2$.

#### Explanation:

Let the nos. be a and b; a,b >0#.

Then, in the Usual Notations,

$A = \frac{a + b}{2} , G = \sqrt{a b} , \mathmr{and} , H = \frac{2 a b}{a + b} \ldots \ldots \ldots \ldots \ldots \left(\star\right)$.

Given Data$\Rightarrow A : G = 5 : 4. \ldots \ldots \ldots . . \left(1\right) , \mathmr{and} , | G - H | = \frac{4}{5.} \ldots \ldots \ldots . \left(2\right)$

Knowing that, $A \ge G \ge H \text{, we rewrite (2) as, } G - H = \frac{4}{5.} \ldots . \left(2 '\right)$.

$\left(1\right) \Rightarrow \frac{a + b}{2 \sqrt{a b}} = \frac{5}{4} \Rightarrow 2 \left(a + b\right) = 5 \sqrt{a b}$

$\Rightarrow 4 \left({a}^{2} + 2 a b + {b}^{2}\right) - 25 a b = 0 \text{, i.e., } 4 {a}^{2} - 17 a b + 4 {b}^{2} = 0$

$\Rightarrow \left(a - 4 b\right) \left(4 a - b\right) = 0 \Rightarrow a = 4 b , \mathmr{and} , b = 4 a$

Case : 1 : a=4b

Then, $G = \sqrt{a b} = 2 b , \mathmr{and} , H = \frac{2 \cdot 4 b \cdot b}{4 b + b} = \frac{8}{5} b$

Hence, by $\left(2 '\right) , 2 b - \frac{8}{5} b = \frac{4}{5} \Rightarrow \frac{2}{5} b = \frac{4}{5} \Rightarrow b = 2$

Thus, in this Case, the Nos. are, $8 , \mathmr{and} , 2$.

Case : 2 : b=4a

We simply notice that the roles of $a \mathmr{and} b$ have interchanged, so,

we immediately jump to the conclusion that, in this Case, the reqd.

Nos. would be $2 , \mathmr{and} , 8$.

Enjoy Maths.!