JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(a_1, a_2, a_3, \ldots\). be a \(GP\) of increasing positive numbers. If the product of fourth and sixth terms is \(9\) and the sum of fifth and seventh terms is \(24 ,\) then \(a_1 a_9+a_2 a_4 a_9+a_5+a_7\) is equal to \(.........\).
- A \(600\)
- B \(606\)
- C \(60\)
- D \(6\)
Answer & Solution
Correct Answer
(C) \(60\)
Step-by-step Solution
Detailed explanation
\(a_4 \cdot a_6=9 \Rightarrow\left(a_5\right)^2=9 \Rightarrow a_5=3\) \(a_5+a_7=24 \Rightarrow a_5+a_5 r^2=24 \Rightarrow\left(1+r^2\right)=8 \Rightarrow r=\sqrt{7}\) \(\Rightarrow a=\frac{3}{49}\) \(\Rightarrow a_1 a_9+a_2 a_4 a_9+a_5+a_7=9+27+3+21=60\)
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