If $x=\frac{2ab}{a+b}$ show that $1/a, 1/x, 1/b$ is an arithmetic progression.

Question

This is the 5th question in my assignment. It's a really freaky looking problem. I haven't come up with an answer yet but I made an attempt with the working out.
Here's the question:
If $x=\frac{2ab}{a+b}$ show that $1/a, 1/x, 1/b$ is a arithmetic progression.
Here's the working out I've done so far:

I stopped there.

Answer 1

First note that

$$\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}$$

or

$$T_1 + T_3 = 2T_2$$ $$T_3 - T_2 = T_2 - T_1$$

Answer 2

If $$\ldots, a_{k - 1}, a_k, a_{k + 1}, \ldots$$ is an arithmetic progression, then $a_k$ is the average of the term before and the term after. In our case: $$\frac{1}{x} = \frac{1}{2} \left(\frac{1}{a} + \frac{1}{b}\right) .$$

Answer 3

You got yourself into trouble when you computed $1/x$. The reciprocal of $x = \dfrac{2ab}{a + b}$ is $$\dfrac{1}{x} = \dfrac{a + b}{2ab}$$

Hence, $$\frac{1}{x} - \frac{1}{a} = \frac{a + b}{2ab} - \frac{1}{a} = \frac{a + b}{2ab} - \frac{2b}{2ab} = \cdots$$ and $$\frac{1}{b} - \frac{1}{x} = \frac{1}{b} - \frac{a + b}{2ab} = \cdots$$