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  • Let  be a G.P. such that <img alt="a_{1}< 0,a_{1}+a_{2}=4" src="http://entrancecorner.oncodecogs.com/gif.l

Let a_{1},a_{2},a_{3},... be a G.P. such that a_{1}< 0,a_{1}+a_{2}=4 and a_{3}+a_{4}=16. If \sum_{i=1}^{9}a_{i}=4\lambda , then \lambda is equal to :
Option: 1 171


Option: 2 \frac{511}{3}


Option: 3 -171


Option: 4 -513
 

Answers (1)

best_answer

Given,

a1 + a2 = 4 \Rightarrowa_1 + a_1r = 4 ………(i)

a3 + a4 = 16 \Rightarrow a_1r^2 + a_1r ^3= 16 …….(ii)

On dividing these

\begin{array}{l}{\frac{1}{r^{2}}=\frac{1}{4} \Rightarrow r^{2}=4} \\\\ {\Rightarrow r=\pm 2} \\\\ {If\,\, r=2,\,\,then\,\,a_{1}(1+2)=4 \Rightarrow a_{1}=\frac{4}{3}}\end{array}

\begin{array}{l}{If\,\,r=-2, \quad a_{1}(1-2)=4 \Rightarrow a_{1}=-4} \\\\ As\, first\, term \,is \,negative,\, so,\,\, r = -2 \\\\ {\sum_{i=1}^{9} a_{i}=\frac{a_{1}\left(r^{9}-1\right)}{r-1}=\frac{(-4)\left((-2)^{9}-1\right)}{-2-1}=\frac{4}{3}(-513)=4 \lambda} \\ {\lambda=-171}\end{array}

Correct Option (3)

Posted by

Ritika Jonwal

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