Search any question & find its solution
Question:
Answered & Verified by Expert
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 A.P. and $|a| < 1,|b| < 1,|c| < 1$, then $x, y, z$ are in
Options:
Solution:
2396 Upvotes
Verified Answer
The correct answer is:
H.P.
H.P.
$$
\begin{array}{ll}
x=\sum_{n=0}^{\infty} a^n=\frac{1}{1-a} & \Rightarrow a=1-\frac{1}{x} \\
y=\sum_{n=0}^{\infty} b^n=\frac{1}{1-b} & \Rightarrow b=1-\frac{1}{y}
\end{array}
$$
$$
\begin{aligned}
& z=\sum_{n=0}^{\infty} c^n=\frac{1}{1-c} \quad \Rightarrow c=1-\frac{1}{z} \\
& a, b, c \text { are in A.P. } \\
& 2 b=a+c \\
& 2\left(1-\frac{1}{y}\right)=1-\frac{1}{x}+1-\frac{1}{y} \\
& \frac{2}{y}=\frac{1}{x}+\frac{1}{z} \\
& \Rightarrow x, y, z \text { are in H.P. }
\end{aligned}
$$
\begin{array}{ll}
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}
\end{array}
$$
$$
\begin{aligned}
& z=\sum_{n=0}^{\infty} c^n=\frac{1}{1-c} \quad \Rightarrow c=1-\frac{1}{z} \\
& a, b, c \text { are in A.P. } \\
& 2 b=a+c \\
& 2\left(1-\frac{1}{y}\right)=1-\frac{1}{x}+1-\frac{1}{y} \\
& \frac{2}{y}=\frac{1}{x}+\frac{1}{z} \\
& \Rightarrow x, y, z \text { are in H.P. }
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.