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?

## Best Answer

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.