NIMCET 2011 Question Paper with solutions

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Question : 29 of 120

Marks: +1, -0

If three positive real numbers

a, b, c(c>a)

are in HP, then

\log (a+c)+\log (a-2 b+c)

is

$2 \log (c-b)$
$2 \log (a+c)$
$2 \log (c-a)$
$\log a+\log b+\log c$

Solution:

Since,

b= {2 a c}/{a+c}

(given)

∴ \log (a+c)+\log (a+c-2 b)

=\log [\{a+c)^{2}-(a+c) 2 b]

=\log [(a+c)^{2}-(a+c) 2 b]

=\log [(a+c)^{2}-4 a c]=\log (c-a)^{2}=2 \log (c-a)

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