The sum of a certain infinite geometric series is 2. The sum of the squares of all the terms is 3. Find the common ratio.

\(\sum \limits_{k=0}^\infty~a r^k = a \dfrac{1}{1-r}=2\\ a = 2(1-r)\\ \sum \limits_{k=0}^\infty~a^2 r^{2k} = \dfrac{a^2}{1-r^2}=3\\ a^2 = 3(1-r^2)\)

\(4(1-r)^2 = 3(1-r^2)\\ 7 r^2-8 r+1\\ \left(r-\dfrac 1 7\right)(r-1)\\ r = \dfrac 1 7 \text{ (1 is an extraneous solution)}\)