If a- b- c are in APand -a- -b- -c- -1 and x-1-a-a2- - y - 1 - b -b2- - z - 1 - c - c2- then x- y- z shall be in

Step1. Sum of infinite terms of Geometric Progression.Geometric Progression for infinite terms is =a/1 r Where, a is first term and r is the common ratio.Given, ...

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Byju's Answer

Standard XII

Mathematics

Harmonic Progression

If a, b, c ar...

Question

If $a, b, c a r e i n A P$ and $| a |, | b |, | c | < 1 a n d x = 1 + a + a^{2} + . . . \infty, y = 1 + b + b^{2} + . . . \infty, z = 1 + c + c^{2} + . . . \infty$ , then $x, y, z$ shall be in

$A . P$

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$G . P$

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$H P$

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$N o n e o f t h e s e$

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Solution

The correct option is C
$H P$

Step1. Sum of infinite terms of Geometric Progression.
Geometric Progression for infinite terms is $= \frac{a}{1 - r}$ Where, $a$ is first term and $r$ is the common ratio.
Given,
$x = 1 + a + a^{2} + \dots \infty x = \frac{1}{(1 - a)}$
$y = 1 + b + b^{2} + \dots \infty y = \frac{1}{(1 - b)}$
$z = 1 + c + c^{2} + \dots \infty z = \frac{1}{(1 - c)}$
Step2. Finding relation between $x, y, z$ ,
Since, $a, b, c$ are in Arithmetic Progression
$1 - a, 1 - b, 1 - c$ are in Arithmetic Progression.
Then, $\frac{1}{1 - a}, \frac{1}{1 - b}, \frac{1}{1 - c}$ are in Harmonic Progression.
So, $x, y, z$ are in Harmonic Progression.
Hence, correct option is $(C)$

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If a,b,careinAPand |a|,|b|,|c|<1andx=1+a+a2+...∞,y=1+b+b2+...∞,z=1+c+c2+...∞, then x,y,z shall be in

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