A) \[-1<x<1\]
B) \[-\infty <x<1\]
C) \[1<x<\infty \]
D) None of these
Correct Answer: C
Solution :
[c] GP = x \[\frac{a}{1-r}=x\] (where, a = 1st term and r = common ratio) \[\Rightarrow \,\,\frac{2}{1-r}=x\] ... (i) (\[\because \] Given \[a=2\] and \[|r|<1\]) \[\Rightarrow \,\,\,-1<r<1\Rightarrow 1>-r>-1\] \[\Rightarrow \,\,1+1>1-r>1-1\] \[\Rightarrow \,\,\,0<1-r<2\] \[\Rightarrow \,\,\,\frac{1}{1-r}>\frac{1}{2},\frac{2}{1-r}>1\] from equation (i) \[x>1\] Hence, \[1<x<\infty .\]You need to login to perform this action.
You will be redirected in
3 sec