Arithmetic Progression- Need help in understanding.

**chakravarthiponmudi** · April 25th 2009, 07:58 PM

Question :If $a^x = b^y = c^z$ and $b^2 = ac$ show that $1/x,1/y,1/z$ are in A.P?
The solution which i got here is as follows.(PFA)
i don't understand how $2/y = 1/x + 1/z$ says it to be A.P.
if i am not wrong, a sequence is said to be in AP if the next number in the sequence differs from the previous number by a constant value D. So I'm confused with the solution.Will someone please explain?

**Soroban** · April 25th 2009, 08:23 PM

Hello, chakravarthiponmudi!

If $a^x = b^y = c^z$ and $b^2 = ac$ show that $\frac{1}{x},\:\frac{1}{y},\:\frac{1}{z}$ are in A.P.

I don't understand how $\frac{2}{y} \:=\: \frac{1}{x} + \frac{1}{z}$ says it to be A.P.

if i am not wrong, a sequence is said to be in AP if the next number in the sequence
differs from the previous number by a constant value $d.$ . . . . You are right!

And they are, too!

If we have: . $\begin{array}{cccc}\dfrac{1}{y} - \dfrac{1}{x} &=& d & [1] \\ \\[-3mm]\dfrac{1}{z} - \dfrac{1}{y} &=& d & [2] \end{array}$

Equate [1] and [2]: . $\frac{1}{y} - \frac{1}{x} \:=\:\frac{1}{z} - \frac{1}{y} \quad\Rightarrow\quad \frac{2}{y} \:=\:\frac{1}{x} + \frac{1}{z}\quad\hdots$ see?

**chakravarthiponmudi** · April 25th 2009, 09:31 PM

Thank you!

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Arithmetic Progression- Need help in understanding.

thanks

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