Question

If $A=1+{{r}^{z}}+{{r}^{2z}}+{{r}^{3z}}+.....\infty $, then the value of $r$ will be
A. \[A{{(1-A)}^{z}}\]
B. \[{{\left( \dfrac{A-1}{A} \right)}^{{}^{1}/{}_{z}}}\]
C. \[{{\left( \dfrac{1}{A}-1 \right)}^{{}^{1}/{}_{z}}}\]
D. \[A{{(1-A)}^{{}^{1}/{}_{z}}}\]

Answer 1

Hint: In this question, we need to find the sum of the infinite terms of the series. Then, from the obtained expression the required value for $r$ is extracted.

Formula used: The geometric sequence is written as
$a,ar,a{{r}^{2}},a{{r}^{3}},...,a{{r}^{n}}$; Here $a$ - first term and $r$ - common ratio
The common ratio is calculated by
$r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
The sum of infinite terms is calculated by
 ${{S}_{\infty }}=\dfrac{a}{1-r}$

Complete step by step solution: The given series is $A=1+{{r}^{z}}+{{r}^{2z}}+{{r}^{3z}}+.....\infty $
From this, we can write
$A=1+({{r}^{z}}+{{r}^{2z}}+{{r}^{3z}}+.....\infty )\text{ }...\text{(1)}$
Where the series $({{r}^{z}}+{{r}^{2z}}+{{r}^{3z}}+.....\infty )$ is a geometric series.
So,
$a={{r}^{z}};$
Common ratio $=\dfrac{{{r}^{2z}}}{{{r}^{z}}}={{r}^{z}}$
Then, the sum of the infinite terms in the obtained series is
$\begin{align}
  & {{S}_{\infty }}=\dfrac{a}{1-r} \\
 & \text{ }=\dfrac{{{r}^{z}}}{1-{{r}^{z}}} \\
\end{align}$
Substituting the obtained sum in equation (1), we get
$\begin{align}
  & A=1+(\dfrac{{{r}^{z}}}{1-{{r}^{z}}}) \\
 & \text{ }=\dfrac{1-{{r}^{z}}+{{r}^{z}}}{1-{{r}^{z}}} \\
 & \text{ }=\dfrac{1}{1-{{r}^{z}}} \\
\end{align}$
So, by simplifying for the required variable, we get
$\begin{align}
  & 1-{{r}^{z}}=\dfrac{1}{A} \\
 & \Rightarrow {{r}^{z}}=1-\dfrac{1}{A} \\
 & \Rightarrow r={{\left( \dfrac{A-1}{A} \right)}^{{}^{1}/{}_{z}}} \\
\end{align}$

Thus, Option (B) is correct.

Note: Here the given series forms a geometric series. By using the appropriate formula for finding the sum of infinite terms, the required value is calculated.