Confusing AP GP

Let ${a_{n}}_{0 \leq n \leq N-1}$ be a geometric progression (GP) with $N$ terms, where the first term is $a_{0}=a$ and the ratio between two consecutive terms is $r$ . The GP can be written as:

$a_{n} = ar^{n}$

Suppose the ratio of the sum of the first 11 terms to the sum of the last 11 terms is $\frac{1}{8}$ . Expressing this:

$\frac{\sum_{k=0}^{10} a_{k}}{\sum_{k=N-11}^{N-1} a_{k}} = \frac{a+ar+ar^{2}+\cdots+ar^{10}}{ar^{N-11}+ar^{N-10}+\cdots+ar^{N-1}} = \frac{1}{r^{N-11}} = \frac{1}{8}$

This gives:

$r^{N-11} = 8$

Furthermore, suppose the ratio of the sum of all terms without the first 9 to the sum of all terms without the last 9 is $2$ . Expressing this:

$\frac{\sum_{k=9}^{N-1} a_{k}}{\sum_{k=0}^{N-10} a_{k}} = \frac{ar^{9}+ar^{10}+\cdots+ar^{N-1}}{a+ar+ar^{2}+\cdots+ar^{N-10}} = r^{9} = 2$

From this:

$r = 2^{\frac{1}{9}}$

Substituting into the equation obtained from the first piece of information:

$2^{\frac{N-11}{9}} = 8$

From this, it is seen that:

$\frac{N-11}{9} = 3$

And thus, $N = 38$ as required.