# Find the values of a and b from arithmetic and geometric series

arithmetic-progressionsgeometric seriessequences-and-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.