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

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{\operatorname{ar}^{ n -1}\left(\left(\frac{1}{ r }\right)^{11}-1\right)}{\left(\frac{1}{ r }-1\right)}}=\frac{1}{8} \Rightarrow \frac{8 a \left( r ^{11}-1\right)}{( r -1)}=\frac{ a \cdot r ^{ n -1}}{ r ^{10}} \cdot \frac{\left(1- r ^{11}\right)}{(1- r )} $
$\Rightarrow 8= r ^{ n -11} \ldots(1)$
Also, $\frac{\frac{\operatorname{ar}^9\left(r^{n-9}-1\right)}{r-1}}{\left(\operatorname{ar}^{n-1}\right)\left(\frac{1}{r^9}\right)\left(\left(\frac{1}{r}\right)^{n-9}-1\right)}=2 \Rightarrow \frac{\operatorname{ar}^9\left(r^{n-9}-1\right)}{r-1}=\left(\frac{2 a \cdot r^{n-1-9}}{r^{n-9}}\right)\left(\frac{1-r^{n-9}}{\frac{1-r}{r}}\right)$
$\therefore r=2^{\frac{1}{9}}$
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