KCET · Maths · Sequences and Series
If \(a_1, a_2 ; a_3, \ldots, a_{10}\) is a geometric progression and \(\frac{a_3}{a_1}=25\), then \(\frac{a_9}{a_5}\) equals
- A \(3\left(5^2\right)\)
- B \(5^4\)
- C \(5^3\)
- D \(2\left(5^2\right)\)
Answer & Solution
Correct Answer
(B) \(5^4\)
Step-by-step Solution
Detailed explanation
Given, GP : \(a_1, a_2, a_3, \ldots, a_{10}\)
and \(\frac{a_3}{a_1}=25\)
\[
\Rightarrow \quad \frac{a r^2}{a}=25 \Rightarrow r^2=25
\]
Now, \(\quad \frac{a_9}{a_5}=\frac{a r^8}{a r^4}=r^4 \cdot=\left(r^2\right)^2=(25)^2=5^4\)
and \(\frac{a_3}{a_1}=25\)
\[
\Rightarrow \quad \frac{a r^2}{a}=25 \Rightarrow r^2=25
\]
Now, \(\quad \frac{a_9}{a_5}=\frac{a r^8}{a r^4}=r^4 \cdot=\left(r^2\right)^2=(25)^2=5^4\)
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