Test Index

UPSC NDA I 2011 Mathematics Paper

© examsnet.com

Question : 15 of 120

Marks: +1, -0

If

\; \frac{1}{b-a} + \; \frac{1}{b-c} = \; \frac{1}{a} + \; \frac{1}{c}

a, b, c

are in

$A P$
GP
$HP$
None of the above

Solution:

Given

\; \frac{1}{b-a} + \; \frac{1}{b-c} = \; \frac{1}{a} + \; \frac{1}{c}

\Rightarrow \; \; \; \frac{1}{b-a} - \; \frac{1}{c} = \; \frac{1}{a} - \; \frac{1}{b-c}

\Rightarrow \; \; \; \frac{c-b+a}{(b-a)c} = \; \frac{b-c-a}{a(b-c)}

\Rightarrow \; \; \; \frac{c-b+a}{(b-a)c} = \; \frac{-(c-b+a)}{a(b-c)}

\Rightarrow \; \; a(b-c) = -(b-a)c

\Rightarrow \; \; a b - a c = -b c + a c

\Rightarrow \; \; b(a+c) = 2 a c

\Rightarrow \; \; b = \; \frac{2 a c}{a+c}

.

Clearly

a, b, c

are in H.P.

© examsnet.com

Go to Question:

Prev Question

Next Question