# Thread: Arithmetic Progression- Need help in understanding.

1. ## Arithmetic Progression- Need help in understanding.

Question :If $\displaystyle a^x = b^y = c^z$ and $\displaystyle b^2 = ac$ show that $\displaystyle 1/x,1/y,1/z$are in A.P?
The solution which i got here is as follows.(PFA)
i don't understand how $\displaystyle 2/y = 1/x + 1/z$ 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?

2. Hello, chakravarthiponmudi!

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

I don't understand how $\displaystyle \frac{2}{y} \:=\: \frac{1}{x} + \frac{1}{z}$ 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 $\displaystyle d.$ . . . . You are right!
And they are, too!

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

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

Thank you!