If a1b+1c,b1c+1a,c1a+1b are in A.P., then

If $a (\frac{1}{b} + \frac{1}{c}), b (\frac{1}{c} + \frac{1}{a}), c (\frac{1}{a} + \frac{1}{b})$ are in A.P., then

A
$a, b, c$ are in A.P.
B
$a, b, c$ are in G.P
C
$a, b, c$ are in H.P.
D
none of these

Solution:

$a (\frac{1}{b} + \frac{1}{c}), b (\frac{1}{c} + \frac{1}{a}), c (\frac{1}{a} + \frac{1}{b})$ are in A.P

$\Rightarrow a (\frac{1}{b} + \frac{1}{c}) + 1, b (\frac{1}{c} + \frac{1}{a}) + 1, c (\frac{1}{a} + \frac{1}{b}) + 1$ are in A.P

A.P

$\Rightarrow a (\frac{b c + c a + a b}{a b c}), b (\frac{b c + c a + a b}{a b c})$

$c (\frac{b c + c a + a b}{a b c})$ are in A.P

$\Rightarrow a, b, c$ are in A.P

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