A) \[3+\frac{3}{2}-\frac{3}{4}+\frac{3}{8}-....\]
B) \[3+\frac{3}{2}-\frac{3}{4}+\frac{3}{8}+....\]
C) \[3-\frac{3}{2}+\frac{3}{4}-\frac{3}{8}+....\]
D) None of these
Correct Answer: C
Solution :
[c] Let first term = a, common ratio = r, where \[-1<r<1\] Then, \[\frac{a}{1-r}=2\] and \[\frac{{{a}^{3}}}{1-{{r}^{3}}}=24\] \[\therefore \,\,\,\frac{1-{{r}^{3}}}{{{(1-r)}^{3}}}=\frac{1}{3}\] i.e \[1-2r+{{r}^{2}}=3(1+r+{{r}^{2}})\] or \[2{{r}^{2}}+5r+2=0\] \[\therefore \,\,\,r=-2\] or \[\frac{-1}{2}\] As \[-1<r<1\] \[\therefore \] we have \[r=-\frac{1}{2}\] \[\therefore \] The series is \[3-\frac{3}{2}+\frac{3}{4}-\frac{3}{8}+...\]
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