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

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

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

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

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