Question

if $x=\displaystyle \sum_{n=0}^\infty \,a^n, $ $y=\displaystyle \sum_{n=0}^\infty \,b^n, $ $z=\displaystyle \sum_{n=0}^\infty \,c^n, $ where $a,b,c$ are in $AP$ and $\left|a\right| <1, \left|b\right|<1, \left|c\right|<1,$ then $x, y,z$ are in :

• A. $HP$
• B. Arithmetico-Geometric Progression
• C. $AP$
• D. $GP$

Answer 1

Answer: (A)

$\because x=\displaystyle \sum_{n=0}^\infty \,a^n=1+a+a^2+...$
$\Rightarrow x=\frac{1}{1-b}$
Similarly,
$y=\frac{1}{1-b}$ and $z=\frac{1}{1-c}$
Now, $a,b, c$ arc in $AP$
$\Rightarrow - a, - b, - c$ are also in $AP$.
$\Rightarrow 1 - a, 1 - b, 1 - c$ are also in $AP$.
$\Rightarrow \frac{1}{1-a}, \frac{1}{1-b}, \frac{1}{1-c}$ are in $HP.$
$\Rightarrow x,y,z$ are in $HP.$
Alternate Solution
$\because x=\frac{1}{1-a}, y=\frac{1}{1-b} , z=\frac{1}{1-c}$
$\Rightarrow a=\frac{x-1}{x}, b=\frac{y-1}{y}, c=\frac{z-1}{z}$
$\because a, b, c$ are in $AP$
$\therefore 2b = a+c$
$\Rightarrow 2\left(\frac{y-1}{y}\right)=\frac{x-1}{x}+\frac{z-1}{z}$
$\Rightarrow 2-\frac{2}{y}=1-\frac{1}{x}+1-\frac{1}{z}$
$\Rightarrow \frac{2}{y}=\frac{1}{x}+\frac{1}{z}$
$\Rightarrow x, y, z$ are in $HP.$