JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(a _1, a _2, a _3, \ldots\) be a \(G.P.\) of increasing positive numbers. Let the sum of its \(6^{\text {th }}\) and \(8^{\text {th }}\) terms be \(2\) and the product of its \(3^{\text {rd }}\) and \(5^{\text {th }}\) terms be \(\frac{1}{9}\). Then \(6\left( a _2+\right.\) \(\left.a_4\right)\left(a_4+a_6\right)\) is equal to
- A \(2 \sqrt{2}\)
- B \(2\)
- C \(3 \sqrt{3}\)
- D \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
\(a r^5+a r^7=2\) \(\left(a r^2\right)\left(a r^4\right)=\frac{1}{9}\) \(a^2 r^6=\frac{1}{9}\) Now, \(r > 0\) \(\operatorname{ar}^5\left(1+r^2\right)=2\) Now, \(ar ^3=\frac{1}{3}\) or \(-\frac{1}{3}\) (rejected) \(r^2=2\) \(r=\sqrt{2}\) \(a=\frac{1}{6 \sqrt{2}}\) Now,…
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