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JEE Mains · Maths · STD 11 - 8. sequence and series
In a increasing geometric series, the sum of the second and the sixth term is \(\frac{25}{2}\) and the product of the third and fifth term is \(25 .\) Then, the sum of \(4^{\text {th }}, 6^{\text {th }}\) and \(8^{\text {th }}\) terms is equal to
- A \(30\)
- B \(26\)
- C \(35\)
- D \(32\)
Answer & Solution
Correct Answer
(C) \(35\)
Step-by-step Solution
Detailed explanation
\(a, ar, ar ^{2}, \ldots\) \(T _{2}+ T _{6}=\frac{25}{2} \Rightarrow \operatorname{ar}\left(1+ r ^{4}\right)=\frac{25}{2}\) \(a^{2} r^{2}\left(1+r^{4}\right)^{2}=\frac{625}{4}\) \(....(1)\) \(T _{3} \cdot T _{5}=25 \Rightarrow\left( ar ^{2}\right)\left( ar ^{4}\right)=25\)…
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