JEE Mains · Maths · STD 11 - 8. sequence and series
Let \({a_1},{a_2}...,{a_{10}}\) be a \(G.P.\) If \(\frac{{{a_3}}}{{{a_1}}} = 25,\) then \(\frac {{{a_9}}}{{{a_{ 5}}}}\) equal
- A \(5^4\)
- B \(4(5^2)\)
- C \(5^3\)
- D \(2(5^2)\)
Answer & Solution
Correct Answer
(A) \(5^4\)
Step-by-step Solution
Detailed explanation
\(\frac{{{a_3}}}{{{a_1}}} = \frac{{{a_1}{r^2}}}{{{a_1}}} = {r^2}\) \( \Rightarrow {r^2} = 25\) Now \(\frac{{{a_9}}}{{{a_5}}} = \frac{{{a_1}{r^8}}}{{{a_1}{r^4}}} = {r^4} = {\left( {25} \right)^2} = {5^4}\)
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