Question

lf $x=\sum _{n=0}{a}^{n}y=\sum _{n=0}{b}^{n}z=\sum _{n=0}{c}^n$ where $a,b,c$ are in A.P. and $|a|<1,|b|<1,|c|<1$, then $x,y,z$ are in

• A. G.P
• B. A.P.
• C. A.G.P
• D. H.P.

Answer 1

Answer: (D)

H.P.

Given

$x=\sum _{n=0}^{\infty }{a}^n=\frac{1}{1-a}\Rightarrow a=1-\frac{1}{x}$
$y=\sum _{n=0}^{\infty }{b}^n=\frac{1}{1-b}\Rightarrow b=1-\frac{1}{y}$
$z=\sum _{n=0}^{\infty }{c}^n=\frac{1}{1-c}\Rightarrow c=1-\frac{1}{z}$

As $a,b,c$ are in $A.P.$
then $2b=a+c$
$\Rightarrow 2(1-\frac{1}{y})=1-\frac{1}{x}+1-\frac{1}{z}$

$\Rightarrow \frac{2}{y}=\frac{1}{x}+\frac{1}{z}$

$⟹\frac{1}{x},\frac{1}{y},\frac{1}{z}$ are in $A.P.$

$\Rightarrow x,y,z$ are in $H.P.$