Review question

Ref: R9177

## Solution

The first term of a geometric progression exceeds the second term by $2$, and the sum of the second and third terms is $\frac{4}{3}$. Calculate the possible values of the first term and of the common ratio of the progression.

The terms of our sequence can be written as $a,ar,ar^2,...$ and the given facts tell us that \begin{align*} a&=ar+2 \\ ar+ar^2&=\frac{4}{3}. \end{align*} Rearranging and factorising, these equations become \begin{align*} a(1-r) &= 2 \\ ar(1+r) &=\frac{4}{3}. \end{align*} We can then divide our second equation by our first giving $\frac{r(1+r)}{1-r}=\frac{2}{3}$ which we can rearrange into a quadratic. \begin{align*} 3r(1+r)&=2(1-r) \\ 3r^2+5r-2&=0 \end{align*}

Using the quadratic formula or by factorisation we find that the common ratio $r=\frac{1}{3}\quad\text{or}\quad-2.$

Now that we have the common difference we can calculate our first term $a=\frac{2}{1-r}=3\quad\text{or}\quad\frac{2}{3}.$

Given further that all the terms of the progression are positive, calculate the sum to infinity.

If all the terms are positive the common ratio cannot be negative, so we must have $r=\frac{1}{3}$ and $a=3$.

Substituting these values into the formula for the sum to infinity we have \begin{align*} S_\infty &= \frac{a}{1-r} \\ &=\frac{3}{2/3}=\frac{9}{2}. \end{align*}