Question

In a geometric progression, the ratio of the sum of first eleven terms to the sum of last eleven terms is $\frac{1}{8}$ and the ratio of the sum of all the terms without first nine terms to the sum of all the terms without last nine terms is 2 , then find the number of terms of the geometric progression.

Answer 1

Answer: 0038

Let $a=$ first term of G.P.
$r=$ common ratio of G.P.
and $n=$ number of terms of G.P.
$\text{ Now, }\frac{\frac{a\left(r^{11}-1\right)}{r-1}}{\frac{{\mathrm{ar}}^{n-1}\left({\left(\frac{1}{r}\right)}^{11}-1\right)}{\left(\frac{1}{r}-1\right)}}=\frac{1}{8}\Rightarrow \frac{8a\left(r^{11}-1\right)}{(r-1)}=\frac{a\cdot r^{n-1}}{r^{10}}\cdot \frac{\left(1-r^{11}\right)}{(1-r)}$
$\Rightarrow 8=r^{n-11}\dots (1)$
Also, $\frac{\frac{{\mathrm{ar}}^9\left(r^{n-9}-1\right)}{r-1}}{\left({\mathrm{ar}}^{n-1}\right)\left(\frac{1}{r^9}\right)\left({\left(\frac{1}{r}\right)}^{n-9}-1\right)}=2\Rightarrow \frac{{\mathrm{ar}}^9\left(r^{n-9}-1\right)}{r-1}=\left(\frac{2a\cdot r^{n-1-9}}{r^{n-9}}\right)\left(\frac{1-r^{n-9}}{\frac{1-r}{r}}\right)$
$\therefore r=2^{\frac{1}{9}}$
So, from equation (1) we get
8=2^{\frac{n-11}{9}} \Rightarrow \frac{n-11}{9}=3 \Rightarrow n-11=27 \Rightarrow n=38