Inspired by Loomba and Hung Woei

Errr...just a correction, my first name is Hung Woei

$\begin{cases}a+a^2+a^3+a^4+\ldots = 2 &\implies\boxed{1}\\ a^2+a^3+a^4+a^5+\ldots = 1 &\implies\boxed{2}\end{cases}$

Now, what we would normally do to find $a$ is to do this:

$\boxed{1}-\boxed{2}:\\ \left(a+a^2+a^3+a^4+\ldots\right)-\left(a^2+a^3+a^4+a^5+\ldots\right) = 2 - 1\\ \implies a=1$

However, this value needs to be verified. We can do this by using the formula for sum of geometric progressions.

Note that $a+a^2+a^3+a^4+\ldots$ is a sum of a geometric progression to infinity, where the first term and common ratio are both $a$ . Therefore,

$a+a^2+a^3+a^4+\ldots=S_{\infty} = \dfrac{a}{1-a} = \dfrac{1}{1-1} = \dfrac{1}{0} \neq 2$

We cannot get back the original value, therefore there is $\boxed{\text{No solution}}$ for $a$

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Alternate method:

Use the formula for sum of geometric progressions to infinity to find the values of $a$ for each equation

$a+a^2+a^3+a^4+\ldots = \dfrac{a}{1-a} = 2\\ a=2(1-a)\\ a=2-2a\\ 3a=2 \implies a=\dfrac{2}{3} \approx 0.667$

$a^2+a^3+a^4+a^5+\ldots = \dfrac{a^2}{1-a} = 1\\ a^2=1-a\\ a^2+a-1 = 0\\ \implies a=\dfrac{-1 \pm \sqrt{5}}{2}$

Since $-1 < a < 1$ , the only solution for the second equation is $a= \dfrac{-1+\sqrt{5}}{2}\approx 0.618$

From here, we see that the two values of $a$ found are different. We can then conclude that there is $\boxed{\text{No solution}}$ for $a$