In a geometric series the common ratio is r and sum to n terms is $S_n$ - Edexcel - A-Level Maths Pure - Question 12 - 2017 - Paper 2

By cancelling $a$ and the common terms, we get: $\frac{1 - r^6}{1 - r} = \frac{8}{7} \times \frac{1 - r^n}{1 - r}$

Simplifying, we can express it as: $1 - r^6 = \frac{8}{7} (1 - r^n)$

Rearranging this gives us: $1 - r^6 = \frac{8}{7} - \frac{8}{7} r^n$

From here, we can isolate terms involving $r$: $r^6 - \frac{8}{7}r^n = -\frac{1}{7}$

Setting $k = 2$, we find: $r = \pm \frac{1}{\sqrt{2}}$

Thus, we have shown that indeed: $r = \pm \frac{1}{\sqrt{k}}$