If the sum of an infinite geometric series is $3$ and the sum of the squares of its terms is $\frac{9}{2}$, find the first term.

Solution

1. Let the series be $a,ar,a{r}^2,\dots$ with $|r|<1$.
2. Sum $$\begin{array}{rl}S& =\frac{a}{1-r}\\ & =3\end{array}$$ and sum of squares $$\begin{array}{rl}{S}^{\prime }& =\frac{a^2}{1-r^2}\\ & =\frac{9}{2}\end{array}$$.
3. Dividing $S^{\prime 2}$ by $S^2$ gives $$\begin{array}{rl}\frac{a^2/(1-r^2)}{(a/(1-r){)}^2}& =\frac{1-r}{1+r}\\ & =\frac{9/2}{9}\\ & =\frac{1}{2}\end{array}$$.
4. $$\begin{array}{rl}2-2r& =1+r⟹3r\\ & =1⟹r\\ & =1/3\end{array}$$.
5. Substituting $r=1/3$ into $a/(1-r)=3$ gives $$\begin{array}{rl}a& =3(2/3)\\ & =2\end{array}$$.
6. Hence the answer is (B).