Iift 2008 Question 37
This is a detailed step-by-step solution for Iift 2008 Question 37. Understand the concept, common mistakes, and expert tips to solve similar questions in IIFT exam.
IIFT 2008
Explanation: It has been given that the terms a, b and c are in harmonic progression.
Therefore, 1/b – 1/a = 1/c – 1/b
2 1 1
b = a + c
2 a + c
b = ac
2ac
b = (a + c) . . . . . (1)
The given expression is log (a + c) + log (a – 2b + c)
log (a + c) + log (a-2b+c) = log ((a + c) (a – ab + c))
Substituting (1), we get,
Log (a + c) + log (a-2b+c) = log ((a + c) (a- (a + c) 4ac + c))
= log (a2 + ac – 4ac + c2 + ac)
= log (a2 + c2 – 2ac)
= log (c – a)2 [since c is greater than a]
= 2 log (c – a)
Therefore, option C is the right answer.