# Proving that three numbers are in A.P.

A number $a$ is the Arithmetic Mean(A.M.) between $b$ and $c$, $b$ is the Geometric Mean(G.M.) between $a$ and $c$. Prove that $$\frac{1}{a}, \frac{1}{c} and \frac{1}{b}$$ are in Arithmetic Progressions(A.P.).

I haven't been able to do much, but this is it :

As, $a$ is the A.M. between $b$ and $c$, $$a = \frac{b+c}{2}$$ or, $$2a = b+c$$

And, $b$ is the G.M. between $a$ and $c$, so, $$b^2 = ac$$ or, $$b = \sqrt{ac}$$How do i proceed next?

• Having you tried proving instead from $\frac{1}{c} - \frac{1}{a} = \frac{1}{b} - \frac{1}{c}$? Commented Mar 9, 2014 at 8:03

As $a$ is the Arithmetic mean, it lies between $b$ and $c$. But, as $b$ is the geometric mean it will lie between $a$ and $c$. This can only mean that $a=b$. And because $a=\frac{b+c}{2}$, $a=b=c$. Clearly,$\frac{1}{a}= \frac{1}{c} = \frac{1}{b}$ and therefore, trivially, in AP.

• I think that the non-trivial case is a spurious solution resulting from squaring Commented Mar 9, 2014 at 8:13
• @Henry oh right. I didn't notice. I am doing a rollback.
– Guy
Commented Mar 9, 2014 at 8:22
• $a=\frac c 4$ yields $b<0$
– alex
Commented Mar 9, 2014 at 8:24
• @alex yeah, it was a spurious solution. I undid my edit.
– Guy
Commented Mar 9, 2014 at 8:25
• @Sabyasachi Sorry, I was too quick...
– alex
Commented Mar 9, 2014 at 8:26

Write $b = 2a - c$ so that $b^2 = (2a - c)^2$. Then with the additional condition $b^2 = ac$, eliminate $b$ and obtain an equation in $a, c$. Factor this equation and consider two cases, one of which is $a = c$.

• wait, there is another case except $a=b=c$? I overlooked something then maybe.
– Guy
Commented Mar 9, 2014 at 8:07
• I only hinted at considering the nontrivial case; I intended the OP to figure out the issues that arise from considering this case. Commented Mar 9, 2014 at 8:10
• I deleted my comment solving the equation.
– Guy
Commented Mar 9, 2014 at 8:11
• No, it's fine. You haven't given away everything. One just has to be careful about the notion of geometric mean. Commented Mar 9, 2014 at 8:11
• This only works if you think that $-2$ is the geometric mean of $1$ and $4$, or that $2$ is the geometric mean of $-1$ and $-4$. I would not accept this, and the reciprocals are then not in arithmetic progression Commented Mar 9, 2014 at 8:12