How do we add the odd-numbered terms of a geometric series?

How do we add the odd-numbered terms of a geometric series?

Question

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 odd-numbered terms (i.e. 1st term + 3rd term + 5th term + …) is $6$. Find the value of $r$.

Suggestion

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 $r$.

We know the terms of the series we are given form a geometric series.

What can we say about the series formed by taking the odd-numbered terms instead?

Solution

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 $\frac{1-r^2}{1-r} =\dfrac{8}{6} \implies 1+r = \dfrac{8}{6} \implies r =\dfrac{1}{3}.$