If x = sum_{n=0}^{infty} a^n,y = sum_{n=0}^{i | vedclass

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(A) Given that $x = \sum_{n=0}^{\infty} a^n = \frac{1}{1-a}$,$y = \sum_{n=0}^{\infty} b^n = \frac{1}{1-b}$,and $z = \sum_{n=0}^{\infty} c^n = \frac{1}

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Harmonic Progression

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(A) Given that $x = \sum_{n=0}^{\infty} a^n = \frac{1}{1-a}$,$y = \sum_{n=0}^{\infty} b^n = \frac{1}{1-b}$,and $z = \sum_{n=0}^{\infty} c^n = \frac{1}{1-c}$.Since $a, b, c$ are in arithmetic progression,we have $2b = a + c$.We need to check if $x, y, z$ are in harmonic progression,which requires $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ to be in arithmetic progression.Note that $\frac{1}{x} = 1-a$,$\frac{1}{y} = 1-b$,and $\frac{1}{z} = 1-c$.Let $A = 1-a$,$B = 1-b$,and $C = 1-c$.Then $2B = 2(1-b) = 2 - 2b = 2 - (a+c) = (1-a) + (1-c) = A + C$.Since $A, B, C$ are in arithmetic progression,their reciprocals $\frac{1}{A}, \frac{1}{B}, \frac{1}{C}$ are in harmonic progression.Thus,$x, y, z$ are in harmonic progression.

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 which progression?

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