(C) Given that $x = 1 + a + a^2 + \dots = \frac{1}{1-a}$,$y = 1 + b + b^2 + \dots = \frac{1}{1-b}$,and $z = 1 + c + c^2 + \dots = \frac{1}{1-c}$ for $

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(C) Given that $x = 1 + a + a^2 + \dots = \frac{1}{1-a}$,$y = 1 + b + b^2 + \dots = \frac{1}{1-b}$,and $z = 1 + c + c^2 + \dots = \frac{1}{1-c}$ for $|a|, |b|, |c| Since $a, b, c$ are in $A.P.$,we have $2b = a + c$.Subtracting each term from $1$,we get $1-a, 1-b, 1-c$ are also in $A.P.$ because $(1-a) + (1-c) = 2 - (a+c) = 2 - 2b = 2(1-b)$.Since $1-a, 1-b, 1-c$ are in $A.P.$,their reciprocals $\frac{1}{1-a}, \frac{1}{1-b}, \frac{1}{1-c}$ must be in $H.P.$Therefore,$x, y, z$ are in $H.P.$

Class 11 Mathematics Sequences and Series Relation between A.P., G.P.

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If $a, b, c$ are in $A.P.$ and $|a|, |b|, |c| < 1$,and $x = 1 + a + a^2 + \dots \infty$,$y = 1 + b + b^2 + \dots \infty$,$z = 1 + c + c^2 + \dots \infty$,then $x, y, z$ shall be in:

AIEEE 2005 All AIEEE

A
$A.P.$
B
$G.P.$
C
$H.P.$
D
None of these

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Relation between A.P., G.P.

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If $a, b, c$ are in $A.P.$ and $|a|, |b|, |c| < 1$,and $x = 1 + a + a^2 + \dots \infty$,$y = 1 + b + b^2 + \dots \infty$,$z = 1 + c + c^2 + \dots \infty$,then $x, y, z$ shall be in:

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