If $a,b,c$ are in GP and $x,y$ are arithmetic mean of $a,b$ and $b,c$ respectively, then $\frac{1}{x} + \frac{1}{y}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(a)
Given $b^{2} = ac,\ \ x = \frac{a + b}{2}$ and $y = \frac{b + c}{2}$
$	herefore\ \frac{1}{x} + \frac{1}{y} = \frac{2}{a + b} + \frac{2}{b + c}$
$= \frac{2(2b + a + c)}{ab + b^{2} + bc + ac}$
$= \frac{2(2b + a + c)}{ab + 2b^{2} + bc}$
$= \frac{2(2b + a + c)}{b(2b + a + c)}$
$= \frac{2}{b}$

Q. If $a,b,c$ are in GP and $x,y$ are arithmetic mean of $a,b$ and $b,c$ respectively, then $\frac{1}{x} + \frac{1}{y}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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