A) \[\frac{1}{4}\]
B) \[\frac{1}{2}\]
C) 2
D) 4
Correct Answer: D
Solution :
Let second term of GP, \[a{{r}^{2-1}}=2\] \[\Rightarrow \] \[ar=2\] ...(i) and sum of infinite terms, \[\frac{a}{1-r}=8\] \[\Rightarrow \] \[a=8(1-r)\] ...(ii) On putting the value from Eq. (ii) in Eq. (i), we get \[8(1-r)r=2\] \[\Rightarrow \] \[4(r-{{r}^{2}})=1\] \[\Rightarrow \] \[4{{r}^{2}}-4r+1=0\] \[\Rightarrow \] \[{{(2r-1)}^{2}}=0\] \[\Rightarrow \] \[r=\frac{1}{2}\] \[\therefore \] \[a=8\left( 1-\frac{1}{2} \right)\] \[\Rightarrow \] \[a=8.\frac{1}{2}\] Hence, first term of the series is 4.
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