Question

Consider an infinite geometric series with first term $a$ and common ratio $r$. If its sum is 4 and the second term is 3/4, then

• A. $a=7/4,r=3/7$
• B. $a=2,r=3/8$
• C. $a=3/2,r=1/2$
• D. $a=3,r=1/4$

Answer 1

Answer: (D)

$a=3,r=1/4$

Given that the sum an infinite geometric series with first term $a$ and common ratio $r$ is 4 and the second term is $\frac{3}{4}$.
That is, $\frac{a}{1-r}=4$ and $ar=\frac{3}{4}$.
Since, $ar=\frac{3}{4}$.
$\Rightarrow r=\frac{3}{4a}$
Now, substitute $r=\frac{3}{4a}$ in $\frac{1}{1-r}=4$; we get
$\frac{a}{1-\frac{3}{4a}}=4$
$\Rightarrow a=4(1-\frac{3}{4a})$
$\Rightarrow a=4-\frac{3}{a}$
$\Rightarrow a=\frac{4a-3}{a}$
$\Rightarrow a^2-4a+3=0$
Now, factoring the trinomial; we get
$a=1$ or $a=3$
When $a=1$, we have $r=\frac{3}{4}$.
When $a=3$, we have $r=\frac{1}{4}$.
Therefore, option D is correct.