Question

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

• A. $HP$
• B. Arithmetice-Geometric Progession
• C. $AP$
• D. $GP$

Answer 1

Answer: (A)

$HP$

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

$x={\sum }_{n=0}^n{a}^n=1+a+a^2+a^3+a^4+....\infty x=\frac{1}{1-a}y={\sum }_{n=0}^n{b}^n=1+b+b^2+b^3+b^4+....\infty y=\frac{1}{1-b}z={\sum }_{n=0}^n{c}^n=1+c+c^2+c^3+c^4+....\infty z=\frac{1}{1-c}x=\frac{1}{1-a}\Rightarrow a=1-\frac{1}{x}y=\frac{1}{1-b}\Rightarrow b=1-\frac{1}{y}z=\frac{1}{1-c}\Rightarrow c=1-\frac{1}{z}$

$a,b,c$ are in AP

$2b=(a+c)2(1-\frac{1}{y})=1-\frac{1}{x}+1-\frac{1}{z}=2-\frac{1}{x}-\frac{1}{z}2-\frac{2}{y}=2-\frac{1}{x}-\frac{1}{z}\frac{2}{y}=\frac{1}{x}+\frac{1}{z}$

$x,y,z$ are inHP