Question

Consider an infinite G.P with first term \[a\] and common ratio \[r\], its sum is 4 and the second term is $\dfrac{3}{4}$, then
A. \[a=\dfrac{7}{4},r=\dfrac{3}{7}\]
B. \[a=\dfrac{3}{2},r=\dfrac{1}{2}\]
C. \[a=2,r=\dfrac{3}{8}\]
D. \[a=3,r=\dfrac{1}{4}\]

Answer 1

Hint: In this question, we are to find the first term and the common ratio of the given series. Here the sum and the second term are given. So, the required common ratio is calculated by using the appropriate formula.

Formula Used:In this question, we are to find the first term and the common ratio of the given series. Here the sum and the second term are given. So, the required common ratio is calculated by using the appropriate formula.

Complete step by step solution:The geometric sequence is written as
$a,ar,a{{r}^{2}},a{{r}^{3}},...,a{{r}^{n}}$
Here $a$ is the first term and $r$ is the common ratio.
The common ratio is calculated by
$r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
The sum of the infinite terms in the G.P series is calculated by
 ${{S}_{\infty }}=\dfrac{a}{1-r}$
Complete Step-by-step Solution:
Consider an infinite geometric series with the first term \[a\] and the common ratio \[r\]be
$a,ar,a{{r}^{2}},a{{r}^{3}},...\infty $
It is given that,
\[{{S}_{\infty }}=4\]
Then,
\[\begin{align}
  & \dfrac{a}{1-r}=4 \\
 & \Rightarrow a=4(1-r)\text{ }...(1) \\
\end{align}\]
Also given that$ar=\dfrac{3}{4}$
Substituting (1),
\[\begin{align}
  & 4(1-r)r=\dfrac{3}{4} \\
 & \Rightarrow 16(r-{{r}^{2}})=3 \\
 & \Rightarrow 16r-16{{r}^{2}}=3 \\
 & \Rightarrow 16{{r}^{2}}-16r+3=0 \\
\end{align}\]
On simplifying,
\[\begin{align}
  & \Rightarrow 16{{r}^{2}}-12r-4r+3=0 \\
 & \Rightarrow 4r(4r-3)-(4r-3)=0 \\
 & \Rightarrow (4r-1)(4r-3)=0 \\
 & \therefore r=\dfrac{1}{4};r=\dfrac{3}{4} \\
\end{align}\]
If the common ratio \[r=\dfrac{1}{4}\], then the first term is
$\begin{align}
  & a=4(1-\dfrac{1}{4}) \\
 & \text{ }=4\left( \dfrac{3}{4} \right) \\
 & \text{ }=3 \\
\end{align}$
If the common ratio \[r=\dfrac{3}{4}\], then the first term is
$\begin{align}
  & a=4(1-\dfrac{3}{4}) \\
 & \text{ }=4\left( \dfrac{1}{4} \right) \\
 & \text{ }=1 \\
\end{align}$
Therefore, the required values are
\[(a,r)=(3,\dfrac{1}{4})\] or \[(1,\dfrac{3}{4})\]

Option ‘D’ is correct

Note: In order to find the first term and the common ratio, the given values (sum and the second term) is to be used within the appropriate formulae.