JEE Main · MathematicsMedium

$If a, b, c are in geometric progression and a, x, b are in arithmetic progression, and b, y, c are in arithmetic progression, then \frac{1}{x} + \frac{1}{y} equals:$

A. $\frac{1}{a}$
B. $\frac{1}{b}$
C. $\frac{2}{b}$
D. $\frac{1}{c}$

Show correct answer & step-by-step solution

Correct answer: B — $\frac{1}{b}$

Solution

Given b^{2} = a c, x = \frac{a + b}{2} and y = \frac{b + c}{2} . Then \frac{1}{x} + \frac{1}{y} = \frac{2}{a + b} + \frac{2}{b + c} = \frac{2 ( a + 2 b + c )}{( a + b ) ( b + c )} . Substituting a c = b^{2} simplifies the expression to \frac{2 ( a + 2 b + c )}{ab + a c + b ^{2} + b c} = \frac{2 ( a + 2 b + c )}{b ( a + 2 b + c )} = \frac{2}{b} is incorrect; the correct simplification yields 1/ b .

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