A) \[\frac{3}{2},\frac{3}{4},\frac{3}{8},\frac{3}{16},.....\]
B) \[\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},.....\]
C) \[\frac{1}{3},\frac{1}{9},\frac{1}{27},\frac{1}{81},.....\]
D) \[1,-\frac{1}{3},\,\frac{1}{{{3}^{2}}},-\frac{1}{{{3}^{3}}},.....\]
Correct Answer: A
Solution :
\[{{({{S}_{1}})}_{\infty }}=\frac{a}{1-r}=3\] or \[a=3\,(\,1-r)\] ?..(i) \[{{({{S}_{2}})}_{\infty }}=\frac{{{a}^{2}}}{1-{{r}^{2}}}=3\] or \[{{a}^{2}}=3\,(1-{{r}^{2}})\] or \[9\,{{(1-r)}^{2}}=3\,(1-{{r}^{2}})\] [by (i)] or \[3\,(1-2r+{{r}^{2}})=1-{{r}^{2}}\] or \[2{{r}^{2}}-3r+1=0\] or \[(r-1)\,(2r-1)=0\], \[\therefore \] \[r=1,\frac{1}{2}\] If \[r=1,\]then \[a=3(1-1)=0\]which is impossible. If \[r=\frac{1}{2},\]then \[a=3\,\left( 1-\frac{1}{2} \right)=3/2\] So first series is 3/2, 3/4, 3/8, 3/16,.....
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