A) 6
B) 3
C) 4
D) 1
Correct Answer: C
Solution :
We have \[ar=2\] and \[{{S}_{\infty }}=8=\frac{a}{1-r}\] \[\Rightarrow \] \[8=\frac{2}{r(1-r)}\left( \because \ a=\frac{2}{r} \right)\] \[\Rightarrow \] \[4r(1-r)=1\Rightarrow 4r-4{{r}^{2}}-1=0\] \[\Rightarrow \] \[4{{r}^{2}}-4r+1=0\Rightarrow \left( r-\frac{1}{2} \right)(4r-2)=0\Rightarrow r=\frac{1}{2}\] So first term\[a=4\].
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