If X = sum_{n=0}^infty a^n,Y = sum_{n=0}^inft | vedclass

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(D) Given $X = \sum_{n=0}^\infty a^n = \frac{1}{1-a}$ for $|a| < 1$.Similarly,$Y = \frac{1}{1-b}$ and $Z = \frac{1}{1-c}$.From these,we have $a = 1 -

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Harmonic

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(D) Given $X = \sum_{n=0}^\infty a^n = \frac{1}{1-a}$ for $|a| Similarly,$Y = \frac{1}{1-b}$ and $Z = \frac{1}{1-c}$.From these,we have $a = 1 - \frac{1}{X}$,$b = 1 - \frac{1}{Y}$,and $c = 1 - \frac{1}{Z}$.Since $a, b, c$ are in arithmetic progression,$2b = a + c$.Substituting the values: $2(1 - \frac{1}{Y}) = (1 - \frac{1}{X}) + (1 - \frac{1}{Z})$.$2 - \frac{2}{Y} = 2 - (\frac{1}{X} + \frac{1}{Z})$.$\frac{2}{Y} = \frac{1}{X} + \frac{1}{Z}$.This implies that $\frac{1}{X}, \frac{1}{Y}, \frac{1}{Z}$ are in arithmetic progression.Therefore,$X, Y, Z$ are in harmonic progression $(H.P.)$.

If $X = \sum_{n=0}^\infty a^n$,$Y = \sum_{n=0}^\infty b^n$,and $Z = \sum_{n=0}^\infty c^n$,where $a, b, c$ are in arithmetic progression and $|a| < 1, |b| < 1, |c| < 1$,then $X, Y, Z$ are in . . . . progression.

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