Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $a$?

share|cite|improve this question
Not sure exactly what you're looking for, but $4a^2+4b^2=25ab$ could be considered a quadratic in $a$, so you could solve for $a$ that way in terms of $b$. – yunone Feb 24 '12 at 5:01
I think the OP hasn't put in efforts in solving. Someone please close the question. – Abhishek Parab Feb 24 '12 at 5:02
@AbhishekParab: Closing the question 10 minutes after it has been posted by a relatively new user (2 days in) is rather harsh. Explaining what to do to improve the question is a rather better course. – Arturo Magidin Feb 24 '12 at 5:04
@Abhishek Parab: You don't know that and we don't close questions for that reason anyway. // anna: The equation should entail $4(a+b)^2=25ab$, which is slightly different from what you have. – anon Feb 24 '12 at 5:06
@Ross: I think you are linking to the wrong question... your link is this question. – Arturo Magidin Feb 24 '12 at 5:37
up vote 6 down vote accepted

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 $$4(2b) = 5\left(\frac{8b}{5}\right).$$

share|cite|improve this answer
Thanks Arturo. I did not think of getting the roots using Discriminant. thank you..:) – vikiiii Feb 24 '12 at 5:20
@anna: You also did your algebra incorrectly, since you went from $2(a+b) = 5\sqrt{ab}$ to $4(a^2+b^2)=25ab$; that's wrong, because $(a+b)^2\neq a^2+b^2$. – Arturo Magidin Feb 24 '12 at 5:21

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 , $t$ $ =$ $\sqrt{\frac{a}{b}}$


A clever person will immediately infer that $t=\frac{1}{2}$

But if its a subjective question we have to justify that also, so


$$t^2 +\frac{1}{t^2} = \frac{17}{4}$$ 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 $a=4b$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.