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If $a, b, c$ are in G.P. and $x, y$ are the arithmetic means between $a, b$ and $b, c$ respectively, then $\frac{a}{x}+\frac{c}{y}$ is equal to
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The correct answer is:
$2$
Given that $a, b, c$ are in G.P.
So, $b^2=a c \ldots(i)$
$x=\frac{a+b}{2} \ldots(ii)$
$y=\frac{b+c}{2} \ldots(iii)$
Now $\frac{a}{x}+\frac{c}{y}=\frac{2 a}{a+b}+\frac{2 c}{b+c}=\frac{2(a b+b c+2 c a)}{a b+a c+b^2+b c}$
$=\frac{2(a b+b c+2 c a)}{(a b+a c+a c+b c)}=2,\left\{b^2=a c\right\}$
Trick : Let $a=1, b=2, c=4$, then obviously $x=\frac{3}{2}$ and $y=3$, then $\frac{1}{3 / 2}+\frac{4}{3}=2$
So, $b^2=a c \ldots(i)$
$x=\frac{a+b}{2} \ldots(ii)$
$y=\frac{b+c}{2} \ldots(iii)$
Now $\frac{a}{x}+\frac{c}{y}=\frac{2 a}{a+b}+\frac{2 c}{b+c}=\frac{2(a b+b c+2 c a)}{a b+a c+b^2+b c}$
$=\frac{2(a b+b c+2 c a)}{(a b+a c+a c+b c)}=2,\left\{b^2=a c\right\}$
Trick : Let $a=1, b=2, c=4$, then obviously $x=\frac{3}{2}$ and $y=3$, then $\frac{1}{3 / 2}+\frac{4}{3}=2$
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