Question

Let $a,b,c$ be in G.P. If $x,y$ are arithmetic means between $a,b$ and $b,c$ respectively, then which of the following is/are correct?

• A. $\frac{1}{x}+\frac{1}{y}=\frac{1}{b}$
• B. $\frac{1}{x}+\frac{1}{y}=\frac{2}{b}$
• C. $\frac{a}{x}+\frac{c}{y}=1$
• D. $\frac{a}{x}+\frac{c}{y}=2$

Answer 1

Answer: (B), (D)

$\frac{a}{x}+\frac{c}{y}=2$

$a,b,c$ are in G.P. $\Rightarrow b^2=ac$
Now, $x=\frac{a+b}{2},y=\frac{b+c}{2}$
$\frac{1}{x}+\frac{1}{y}=\frac{2}{a+b}+\frac{2}{b+c}=2\left[\frac{b+c+a+b}{(a+b)(b+c)}\right]=2\left[\frac{2b+a+c}{ab+ac+b^2+bc}\right]$
$=2\left[\frac{2b+a+c}{2b^2+ab+bc}\right](\because ac=b^2)$
$=\frac{2}{b}\frac{(2b+a+c)}{(2b+a+c)}=\frac{2}{b}$

Now,
$\frac{a}{x}+\frac{c}{y}=\frac{2a}{a+b}+\frac{2c}{b+c}$
$=2\left[\frac{ab+ac+ac+bc}{2b^2+ab+bc}\right](\because ac=b^2)$
$=2\left[\frac{ab+bc+2b^2}{2b^2+ab+bc}\right]$
$=2$