Indian Institute of Foreign Trade 2008 Solved Paper

Question : 37 of 150

Marks: +1, -0

If three positive real numbers a, b and c (c > a) are in Harmonic Progression, then

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

is equal to:

Solution:

It has been given that the terms and are in harmonic progression
Therefore,

1/b-1/a=1/c-1/b

2/b=1/a+1/c

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

b={2ac}/(a+c)

.....(1)
The given expression is

\log+\log(a−2b+c)

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

Substituting (1), we get,

\log(a+c)+\log(a-2b+c)=\log((a+c)-(a-{4ac}/(a+c)+c))

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

=\log(a^{2}+c^{2}-2ac)

=\log(c-a)^{2}

[Since c is greater than a]

=2\log(c-a)

Therefore, option C is the right answer.

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