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VITEEE 2012 Paper

Section: Mathematics

Question : 90 of 120

Marks: +1, -0

a^2

b^2

c^2

then x, y and z are in

Solution:

x = 1 + a +

a^2

+ ... ∞ =

\frac{1}{1-a}

y = 1 + b +

b^2

+ ... ∞ =

\frac{1}{1-b}

and z = 1 + c +

c^2

+ ... ∞ =

\frac{1}{1-c}

Since, a, b and c are in AP.

⇒ 1 - a, 1 - b and 1 - c are also in AP.

⇒

\frac{1}{1-a} \cdot \frac{1}{1-b}

and

\frac{1}{1-c}

are in HP.

∴ x, y and z are in HP.

Note that if the common ratio of a GP is not less than 1, then we do not determined the sum of an infinite GP that series

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