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

Let \(a_{1}, a_{2}, a_{3}, \ldots\) be a G.P. such that \(a_{1}<0\); \(a_{1}+a_{2}=4\) and \(a_{3}+a_{4}=16 .\) If \(\sum\limits_{i=1}^{9} a_{i}=4 \lambda,\) then \(\lambda\) is equal to 

  1. A \(-171\)
  2. B \(171\)
  3. C \(\frac{511}{3}\)
  4. D \(-513\)
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Answer & Solution

Correct Answer

(A) \(-171\)

Step-by-step Solution

Detailed explanation

\(a_{1}+a_{2}=4\) \(\mathrm{r}^{2} \mathrm{a}_{1}+\mathrm{r}^{2} \mathrm{a}_{2}=16\) \(\Rightarrow \mathrm{r}^{2}=4 \Rightarrow \mathrm{r}=-2 \quad\) as \(\mathrm{a}_{1}<0\) and \(a_{1}+a_{2}=4\) \(a_{1}+a_{1}(-2)=4 \Rightarrow a_{1}=-4\)…
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