SATHEE: Sequences and Series 3 Question 4

Sequences and Series 3 Question 4

4.

Let $a_{1}, a_{2}, \dots, a_{10}$ 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) $2 (5^{2})$

(c) $4 (5^{2})$

(d) $5^{4}$

Show Answer

Answer:

Correct Answer: 4. (d)

Solution:

Let $r$ be the common ratio of given GP, then we have the following sequence $a_{1}, a_{2} = a_{1} r, a_{3} = a_{1} r^{2}, \dots, a_{10} = a_{1} r^{9}$

Now, $a_{3} = 25 a_{1}$

$\begin{aligned} \Rightarrow & a_{1} r^{2} & = 25 a_{1} \\ \Rightarrow & r^{2} & = 25 \end{aligned}$

Consider, $\frac{a_{9}}{a_{5}} = \frac{a_{1} r^{8}}{a_{1} r^{4}} = r^{4} = (25)^{2} = 5^{4}$

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