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

Solution

1. Let the infinite geometric series be $a,ar,a{r}^2,\dots$, where the sum is $\frac{a}{1-r}=3$ and the sum of the squares is $\frac{a^2}{1-r^2}=\frac{9}{2}$.
2. Square the first equation to get $\frac{a^2}{(1-r{)}^2}=9$, then divide this by the second equation to obtain $$\begin{array}{rl}\frac{1-r^2}{(1-r{)}^2}& =\frac{9}{9/2}\\ & =2\end{array}$$.
3. Simplify the expression to $$\begin{array}{rl}\frac{(1-r)(1+r)}{(1-r{)}^2}& =\frac{1+r}{1-r}\\ & =2\end{array}$$, which leads to $1+r = 2-2r$, so $3r = 1$ and $r=\frac{1}{3}$.
4. Substitute $r=\frac{1}{3}$ back into the first equation: $\frac{a}{1-1/3}=3$, which simplifies to $\frac{a}{2/3}=3$, giving $a=2$.
5. Hence the answer is (B).