Find the value of a? Let $a$ and $b$ be  two positive  numbers such that $a\gt b$. Let $G$ be the geometric mean of $a$ and $b$ (that is, $G=\sqrt{ab}$), and $H$ be the Harmonic mean of $a$ and $b$, that is, $$H = \frac{2}{\frac{1}{a}+\frac{1}{b}} = \frac{2ab}{a+b}.$$
If $4G = 5H$, what is the value of $a$?
 A: The Harmonic Mean of $a$ and $b$ is
$$\frac{2}{\frac{1}{a}+\frac{1}{b}} = \frac{2ab}{a+b}.$$
The Geometric Mean of $a$ and $b$ is
$$\sqrt{ab}.$$
So, to state the problem you have in a way that would be actually intelligible would be:

Let $a$ and $b$ be positive numbers such that $a\gt b$; assume that
  $$4\times\text{geometric-mean(a,b)} = 4\sqrt{ab} = 5\left(\frac{2ab}{a+b}\right) = 5\times\text{harmonic-mean}(a,b).$$
  What is the value of $a$?

We have
$$\begin{align*}
4\sqrt{ab} &= \frac{10ab}{a+b}\\
4(a+b) &= \frac{10ab}{\sqrt{ab}}\\
2(a+b) &= 5\sqrt{ab}\\
4(a+b)^2 &= 25ab\\
4a^2 + 8ab + 4b^2 &= 25ab\\
4a^2 -17ab + 4b^2 &=0.
\end{align*}$$
You can view this as a quadratic equation in $a$; the solutions are given by
$$\frac{17b - \sqrt{(17b)^2 - 64b^2}}{8} = \frac{17b-\sqrt{225b^2}}{8} = \frac{17b-15b}{8} = \frac{b}{4}$$
(which is impossible since $a\gt b$) and
$$\frac{17b + \sqrt{(17b)^2 - 64b^2}}{8} = \frac{17b + \sqrt{225b^2}}{8} = \frac{32b}{8} = 4b.$$
So the answer is that $a$ must be $4b$.
You can verify this works: the Geometric Mean of $b$ and $4b$ is $\sqrt{4b^2} = 2b$; the Harmonic mean is 
$$\frac{2(4b)b}{4b+b} = \frac{8b^2}{5b} = \frac{8b}{5}.$$
And 
$$4(2b) = 5\left(\frac{8b}{5}\right).$$
A: According to the given information we have,
$$4\sqrt{ab} = 5(\frac{2ab}{a+b})$$
$$(a+b) =\frac{5}{2}\sqrt{ab})$$
$$\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}} = \frac{5}{2}$$
Let , $t$ $ =$ $\sqrt{\frac{a}{b}}$
$$[t+\frac{1}{t}=\frac{5}{2}].....................Eq(1)$$
A clever person will immediately infer that $t=\frac{1}{2}$
But if its a subjective question we have to justify that also, so
$$({t+\frac{1}{t}})^2=\frac{25}{4}$$
$$t^2 +\frac{1}{t^2} = \frac{17}{4}$$
NOW,
$$(t-\frac{1}{t})^2= t^2+\frac{1}{t^2} -2 =\frac{9}{4}$$
$$[t-\frac{1}{t}=\frac{3}{2}   ]   .....................Eq(2)$$
neglecting the negative value as we know that$ L.H.S.>0$ , since $,t>0$
From Eq(1) and Eq(2) we have $t=4$, hence $a=4b$
