A geometric series has first term \(a\) and common ratio \(r\), where \(|r|<1\). The sum to infinity of the series is \(8\). The sum to infinity of the series obtained by adding all the off-numbered terms (i.e. 1st term + 3rd term + 5th term + …) is \(6\). Find the value of \(r\).

If \(|r|<1\), we know that the sum to infinity of the GP (geometric progression) $a, ar, ar^2, … $ is given by

\[S_\infty=\frac{a}{1-r}.\]

Hence in our case we have that \[\dfrac{a}{1-r}=8.\]

The odd-numbered terms of the above series are \(a, ar^2, ar^4, ....\) which themselves form a GP with first term \(a\), common ratio \(r^2\).

We have \(|r|<1 \iff |r^2|<1\), and so this series also has a sum to infinity, \(T_\infty\), where \[T_\infty =\frac{a}{1-r^2} =6. \]

Dividing \(\dfrac{a}{1-r}=8\) by \(\dfrac{a}{1-r^2} =6\) gives us \[\frac{1-r^2}{1-r} =\dfrac{8}{6} \implies 1+r = \dfrac{8}{6} \implies r =\dfrac{1}{3}.\]