Question

if $x=\sum _{n=0}^{\infty }{a}^n,$ $y=\sum _{n=0}^{\infty }{b}^n,$ $z=\sum _{n=0}^{\infty }{c}^n,$ 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)

$\because x=\sum _{n=0}^{\infty }{a}^n=1+a+a^2+...$
$\Rightarrow x=\frac{1}{1-b}$
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 $HP.$