Sequences and Series 3 Question 4
4. Let $a _1, a _2, \ldots, a _{10}$ be a GP. If $\frac{a _3}{a _1}=25$, then $\frac{a _9}{a _5}$ equals
(2019 Main, 11 Jan I)
(a) $5^{3}$
(b) $2\left(5^{2}\right)$
(c) $4\left(5^{2}\right)$
(d) $5^{4}$
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Answer:
Correct Answer: 4. (b)
Solution:
- Let $r$ be the common ratio of given GP, then we have the following sequence $a _1, a _2=a _1 r, a _3=a _1 r^{2}, \ldots, a _{10}=a _1 r^{9}$ Now,
$$ a _3=25 a _1 $$
$$ \begin{aligned} \Rightarrow & & a _1 r^{2} & =25 a _1 \\ \Rightarrow & & r^{2} & =25 \end{aligned} $$
Consider, $\frac{a _9}{a _5}=\frac{a _1 r^{8}}{a _1 r^{4}}=r^{4}=(25)^{2}=5^{4}$
