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JEE Mains · Maths · STD 11 - 8. sequence and series

Let \(A _{1}, A _{2}, A _{3}, \ldots \ldots\) be an increasing geometric progression of positive real numbers. If \(A _{1} A _{3} A _{5} A _{7}=\frac{1}{1296}\) and \(A _{2}+ A _{4}=\frac{7}{36}\), then, the value of \(A _{6}+ A _{8}+ A _{10}\) is equal to

  1. A \(33\)
  2. B \(37\)
  3. C \(43\)
  4. D \(47\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(43\)

Step-by-step Solution

Detailed explanation

\(A _{1} \cdot A _{3} \cdot A _{5} \cdot A _{7}=\frac{1}{1296}\) \(\left( A _{4}\right)^{4}=\frac{1}{1296}\) \(A _{4}=\frac{1}{6}.....(1)\) \(A _{2}+ A _{4}=\frac{7}{36}\) \(A _{2}=\frac{1}{36}.....(2)\) \(A _{6}=1\) \(A _{8}=6\) \(A _{10}=36\) \(A _{6}+ A _{8}+ A _{10}=43\)
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