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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Solution:

  1. 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 $a=3, r=1 / 4$



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