In a G.P the ratio of the sum of the first eleven terms to the sum of the last eleven terms is $\frac{1}{8}$ and the ratio of the sum of all the terms without the first nine to the sum of all the terms without last nine is 2. The number of terms of the G.P is less than

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Detailed Solution

${a,ar}_{1} ... {ar}^{n-1}$

$\begin{array}{l} \to \frac{\frac{a (r^{11} - 1)}{r - 1}}{\frac{{ar}^{n - 1} ({(\frac{1}{r})}^{11} - 1)}{\frac{1}{r} - 1}} = \frac{1}{8} \\ \to \frac{8 a (r^{11} - 1)}{r - 1} = \frac{{ar}^{n - 1} (1 - r^{11})}{r^{10} (1 - r)} \\ \to 8 = 1^{n - 11} - - - (1) \\ \frac{\frac{{ar}^{9} (r^{n - 9} - 1)}{r - 1}}{\frac{a r^{n - 1} (1 / r)^{9} ({(\frac{1}{r})}^{n - 9} - 1)}{(\frac{1}{r} - 1)}} = 2 \\ \Rightarrow \frac{a r^{9} (r^{n - 9} - 1)}{r - 1} = \frac{2 \cdot a r^{n - 1 - 9}}{r^{n - 9}} \frac{(1 - r^{n - 9})}{\frac{1 - r}{r}} \end{array}$

$\begin{array}{l} \Rightarrow r^{9} = \frac{2}{r} \times r \times r \Rightarrow r = 2^{1.9} - \dots - - (2) \\ From 1 and 2 \Rightarrow 8 = 2^{\frac{n - 11}{9}} \\ \Rightarrow \frac{n - 11}{9} = 3 \\ \Rightarrow n - 11 = 27 \\ \Rightarrow n = 38 \end{array}$