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Sequence and Series

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Question : 37 of 86

Marks: +1, -0

If

a, b, c

., a, x, b

b, y, c

are both in A.P., the value of

\;\frac{1}{x}+\;\frac{1}{y}

is

$\frac{1}{b}$
$\frac{2}{b}$
$-\frac{2}{b}$
None

Solution:

Concept:

Use GP condition

b^2 = ac

and AP conditions to express

x

and

y

, then simplify

\frac{1}{x} + \frac{1}{y}

.

Explanation:

Given

a,b,c

in GP

\Rightarrow b^2 = ac

.

a,x,b

in AP

\Rightarrow 2x = a+b \Rightarrow x = \frac{a+b}{2}

.

b,y,c

in AP

\Rightarrow 2y = b+c \Rightarrow y = \frac{b+c}{2}

.

Compute

\frac{1}{x} + \frac{1}{y} = \frac{2}{a+b} + \frac{2}{b+c} = 2\left(\frac{1}{a+b} + \frac{1}{b+c}\right)

.

= 2\cdot\frac{(b+c)+(a+b)}{(a+b)(b+c)} = 2\cdot\frac{a+2b+c}{(a+b)(b+c)}

.

Now

(a+b)(b+c) = ab + ac + b^2 + bc = ab + b^2 + bc + ac

(since

ac = b^2

).

= ab + b^2 + bc + b^2 = b(a + b + c + b) = b(a+2b+c)

.

Thus

\frac{1}{x} + \frac{1}{y} = 2\cdot\frac{a+2b+c}{b(a+2b+c)} = \frac{2}{b}

(assuming

a+2b+c \neq 0

).

Answer:

\frac{2}{b}

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