Sequences and Series 4 Question 8
8.
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
(2000, 2M)
(a) $a=4 / 7, r=3 / 7$
(b) $a=2, r=3 / 8$
(c) $a=3 / 2, r=1 / 2$
(d) $a=3, r=1 / 4$
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Answer:
Correct Answer: 8. (d)
Solution:
- Since, sum $=4$ and second term $=\frac{3}{4}$.
It is given first term $a$ and common ratio $r$.
$ \begin{aligned} & \Rightarrow \quad \frac{a}{1-r}=4, \text { ar }=\frac{3}{4} \\ & \Rightarrow \quad r=\frac{3}{4 a} \\ & \Rightarrow \quad \frac{a}{1-\frac{3}{4 a}}=4 \\ & \Rightarrow \quad \frac{4 a^{2}}{4 a-3}=4 \\ & \Rightarrow \quad(a-1)(a-3)=0 \\ & \Rightarrow \quad a=1 \quad \text { or } 3 \end{aligned} $
When $a=1, r=3 / 4$
and when $a=3, r=1 / 4$
