$\left[\begin{array}{l}{[\text { [geometric progression }]} \\ {[\text { Evenness and oddness }]}\end{array}\right.$

Can a geometric progression be extracted from the geometric progression $1,1 / 2,1 / 4,1 / 8, \ldots$ with the sum of its terms equal to a) $1 / 7$; b) $1 / 5$?
In the highlighted progression, the first term is of the form $2^{-k}$, and the common ratio is $-2^{-p}$, where $k \geq 0, p>0$ are integers. Therefore, its sum $S=\frac{2^{-k}}{1-2^{-p}}=\frac{2^{p-k}}{2^{p}-1}$.

a) Let $p=k=3$. Then $S=1 / 7$.

b) For the sum $S$ to equal ${ }^{1} / 7$, it is necessary that $2^{p}-1=5 \cdot 2^{p-k}$. However, $2^{p}-1$ is divisible by 5 only when $p$ is a multiple of 4, and in that case, $2^{p}-1$ is divisible by $16-1=15$.

## Answer

a) Possible; b) not possible.