Question

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

Answer 1

Hint: We will use the sum of infinite series to find the equation in $a$ and $r$. Next, use the given term of the geometric series to find a relation in $a$ and $r$. Substitute the value of $a$ in terms of $r$ in the first equation and solve it to find the value of $r$. Put back the value of $r$ in the relationships formed to calculate the value of $a$.

Complete step by step answer:

We are given that the sum of infinite terms of GP is 4.
The first term of the GP is $a$ and the common ratio is $r$.
Then, the geometric series is of the type $a,ar,a{r^2},a{r^3}.......$
As it is known that the sum of the infinite sum of series of a GP is $\dfrac{a}{{1 - r}}$, where $a$ is the first term of the geometric progression and $r$ is the common ratio of GP.
This implies $\dfrac{a}{{1 - r}} = 4$........eqn. (1)
Also, it is given that the second term of the GP is 4.
That implies, \[ar = \dfrac{3}{4}\]......eqn. (2)
From here we can calculate the value of $a$ is $\dfrac{3}{4r}$.
We will now substitute the value of $a$ in equation (1) and we will find the values of $r$.
$\dfrac{{\dfrac{{3}}{4r}}}{{1 - r}} = 4$
We will cross multiply and rearrange the equation as,
\[16{r^2} - 16r + 3 = 0\]
On factorising the above equation we will get,
\[\left( {4r - 3} \right)\left( {4r - 1} \right) = 0\]
Put each factor equal to 0 to find the value of therefore, we have
\[r = \dfrac{3}{4},\dfrac{1}{4}\]
We will substitute $r = \dfrac{1}{4}$ in the equation. 2 to find the value of $a$.
\[a = \dfrac{{\dfrac{3}{4}}}{{\dfrac{1}{4}}} = 3\]
When $r = \dfrac{3}{4}$ we have
\[a = \dfrac{{\dfrac{3}{4}}}{{\dfrac{3}{4}}} = 1\]
Hence, option C is correct.

Note: If the first term of the series is a and the common ratio is $r$ then the corresponding geometric series is written as $a,ar,a{r^2},a{r^3}.......$ When the series is infinite and the common ratio $r$ is such that $\left| r \right| < 1$, then the sum of the series is $\dfrac{a}{{1 - r}}$.