A) 1
B) 3
C) 7
D) None of these
Correct Answer: A
Solution :
Given that \[\frac{a({{r}^{n}}-1)}{r-1}=255\] \[(\because \ \ r>1)\] ?..(i) \[a{{r}^{n-1}}=128\] ?..(ii) and common ratio \[r=2\] ?..(iii) From (iii), (i) and (ii) we get \[{{a}_{1}}={{h}_{1}}=2,\ {{a}_{10}}={{h}_{10}}=3\] ?..(iv) and\[\frac{a({{2}^{n}}-1)}{2-1}=255\] .....(v) Dividing (v) by (iv) we get \[\frac{{{2}^{n}}-1}{{{2}^{n-1}}}=\frac{255}{128}\]\[\Rightarrow \]\[2-{{2}^{-n+1}}=\frac{255}{128}\] \[\Rightarrow \]\[{{2}^{-n}}={{2}^{-8}}\]\[\Rightarrow \]\[n=8\] Putting \[n=8\] in equation (iv), we have \[a\ .\ {{2}^{7}}=128={{2}^{7}}\]or \[a=1\].
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