A) 5
B) 1
C) 4
D) 3
Correct Answer: C
Solution :
[c] Let the G.P be \[a,ar,a{{r}^{2}},......\] \[S=a+ar+a{{r}^{2}}+.......+to\,\,2n\,\,tern=\frac{a({{r}^{2n}}-1)}{r-1}\] The series formed by taking term occupying odd places is \[{{S}_{1}}=a+a{{r}^{2}}+a{{r}^{4}}+........\] to n terms \[{{S}_{1}}=\frac{a\left[ {{({{r}^{2}})}^{n}}-1 \right]}{{{r}^{2}}-1}\Rightarrow {{S}_{1}}=\frac{a({{r}^{2n}}-1)}{{{r}^{2}}-1}\] Now, \[S=5{{S}_{1}}\] or \[\frac{a({{r}^{2n}}-1)}{r-1}=5\frac{a({{r}^{2n}}-1)}{{{r}^{2}}-1}\] \[\Rightarrow 1=\frac{5}{r+1}\Rightarrow r+1=5\therefore r=4\]
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