Let $a,b,c$ be in AP and $\left| a \right| < 1,\;\left| b \right| < 1,\;\left| c \right| < 1$. If $x = 1 + a + {a^2} + ...{\rm{to}}\;\infty $, $y = 1 + b + {b^2} + ...{\rm{to}}\;\infty $ and $z = 1 + c + {c^2} + ...\infty $, then $x,\;y,\;z$ are in

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

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Analyzing +4 | -1

Let

a,b,c

be in AP and

\left| a \right| < 1,\;\left| b \right| < 1,\;\left| c \right| < 1

. If

x = 1 + a + {a^2} + ...\mathrm{to}\;\infty

y = 1 + b + {b^2} + ...\mathrm{to}\;\infty

and

z = 1 + c + {c^2} + ...\infty

, then

x,\;y,\;z

are in

View answer and explanation

Correct option: C

Option C (HP) is correct because if a, b, c are in AP, then 1/(1-a), 1/(1-b), 1/(1-c) are in HP, which corresponds to x, y, and z.

Practice All Chapter MCQs →

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\sum\limits_{{i=0}}^{2012} a_i

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A P

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G P

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\frac{3}{4}