A) 63
B) 189
C) 728
D) 364
Correct Answer: C
Solution :
Let \[a,ar,a{{r}^{2}}a{{r}^{3}},a{{r}^{4}},a{{r}^{5}}\]be six terms of a G.P. where 'a' is first term and r is common ratio. According to given conditions, we have \[a{{r}^{3}}-a=5\Rightarrowa({{r}^{3}}-1)=52\] ?(1) and\[a+ar+a{{r}^{2}}=26\]\[\Rightarrow \]\[a(1+r+{{r}^{2}})=26\]?(2) To find\[a(1+r+{{r}^{2}}+{{r}^{3}}+{{r}^{4}}+{{r}^{5}})\]Consider \[a[1+r+{{r}^{2}}+{{r}^{3}}+{{r}^{4}}+{{r}^{5}}]\] \[=a[1+r+{{r}^{2}}+{{r}^{3}}(1+r+{{r}^{2}})]\] \[=a[1+r+{{r}^{2}}][1+{{r}^{3}}]\] ?(3) Divide (1) by (2), we get\[\frac{{{r}^{3}}-1}{1+r+{{r}^{2}}}=2,\] We know\[{{r}^{3}}-1=(r-1)(1+r+{{r}^{2}})\] \[\therefore \]\[r-1=2\Rightarrow r=3\]and\[a=2\] \[\therefore \]\[a(1+r+{{r}^{2}}+{{r}^{3}}+{{r}^{4}}+{{r}^{5}})\] \[=a(1+r+{{r}^{2}})(1+{{r}^{3}})\] \[=2(1+3+9)(1+27)\] \[=26\times 28=728\]
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