Find the values of a and b from arithmetic and geometric series The $1^{st}$ , $2^{nd}$ and $3^{rd}$ terms of an arithmetic series are $a, b, a^2$, where $a$ is a negative number. The $1^{st}$, $2^{nd}$ and $3^{rd}$ terms of a geometric series are $a, a^2,b$.
Find the values of $a$ and $b$ and find the sum to infinity of the geometric series.
What I have tried:
By equating the geometric series to the following:
$a^2=ar, b=ar^2 \implies r=a$
Then equating for the arithmetic series.
$a^3-a = a^2-a^3 \implies 2a^3-a^2-a=0 \implies a(a-1)(2a+1) \\ a=0, a=1, a=-\frac{1}{2} \\
\implies b=0, b=1, b=-\frac{1}{8}$
After plugging in the values of $a$ for $b=ar^2$ when $r=a$.
This produces the following series:
$$(0,0) = 0,0,0 ... \\ (1,1) = 1, 1, 1, ... \\ (-\frac{1}{2},-\frac{1}{8})=-\frac{1}{2},\frac{1}{4},-\frac{1}{8}...$$
The first series is equal to zero, the second is equal to zero, and the third has the following geometric series:
$$\sum_{n=0}^{\infty}(-1)^n \left(\frac{1}{2}\right)^n =\frac{2}{3}$$
Are my calculations correct for this?

Looks mostly OK – note that $a$ is negative, so $a=-\frac 1 2$, and $b=-\frac 1 8$ (you don't need the other two cases).
The sum is $\frac{-\frac 1 2}{1 + \frac 1 2} = -\frac 1 3$, though. The formula for a geometric series is $\frac{a}{1 - r}$, and here, $r=a$. Your sum should be from $n=1$; as it stands, it includes a term of $1$ at the beginning, which is why your answer is too large by $1$.
One other thing – the sum of the series $1,1,1,\dots$ is not $0$ – it is undefined.
