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. If \(\mathrm{a}_3 \mathrm{a}_5=729\) and \(\mathrm{a}_2+\mathrm{a}_4=\frac{111}{4}\), then \(24\left(a_1+a_2+a_3\right)\) is equal to
- A \(131\)
- B \(130\)
- C \(129\)
- D \(128\)
Answer & Solution
Correct Answer
(C) \(129\)
Step-by-step Solution
Detailed explanation
Let the \(\mathrm{I}^{\text {st }}\) term of G.P. be a \& common ratio be r \(\begin{aligned} \mathrm{a}_3 \mathrm{a}_5 & =\operatorname{ar}^2 \cdot \operatorname{ar}^4=729 \\ & =\mathrm{a}^2 \mathrm{r}^6=729 \\ & =\mathrm{ar}^3=27 \quad ....(i) \end{aligned}\)…
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