Check for false roots too!

Let the $n$ th term of the original progression be $a_{n} = ar^{n-1}$ for $|r| \lt 1$ . We are then given that

$\dfrac{a}{1 - r} = 1 \Longrightarrow a = 1 - r$ .

The $n$ th term of the "cubed" progression will be $a^{3}(r^{3})^{n-1}$ , and thus we are given that

$\dfrac{a^{3}}{1 - r^{3}} = 3 \Longrightarrow (1 - r)^{3} = 3(1 - r)(1 + r + r^{2})$ ,

where we substituted in $a = 1 - r$ . Canceling out $(1 - r)$ from both sides and simplifying, we find that

$1 - 2r + r^{2} = 3 + 3r + 3r^{2} \Longrightarrow 2r^{2} + 5r + 2 = 0 \Longrightarrow (2r + 1)(r + 2) = 0$ .

Now as we require that $|r| \lt 1$ we must have that $r = -\dfrac{1}{2}$ , in which case $a = 1 - r = \dfrac{3}{2}$ .

The $n$ th term of the "squared" progression will be $a^{2}(r^{2})^{n-1}$ , and so the sum of this progression will be

$\dfrac{a^{2}}{1 - r^{2}} = \dfrac{\dfrac{9}{4}}{1 - \dfrac{1}{4}} = \boxed{3}$ .

Let's denote the first term as $a$ and the ratio as $q$ . We have that:

$\frac{a}{1-q} = 1$

$\fbox{a = 1-q}$

$\frac{a^3}{1 - q^3} = 3$

Expanding $(1-q)^3$ and plugging the value of $a$ :

$\frac{(1-q)^3}{(1-q)(1+q+q^2)} = 3$

$\frac{(1-q)^2}{1+q+q^2} = 3$

$2q^2 + 5q + 2= 0$

$q = -2$ or $q = -0.5$ . Since the series is convergent, $|q| < 1$ and, so, $\fbox{q = -0.5}$ and $\fbox{a = 1.5}$

We're looking for:

$\frac{a^2}{1-q~2} = \frac{1.5^2}{1 - (-0.5)^2} = \fbox{3}$