# The sum of the first two terms of a geometric sequence is 8/9 and of the first three terms is 26/27. What is the first term, common ratio and sum to infinity ?

Aug 14, 2016

The first term can be $\frac{2}{3}$ with common ratio $\frac{1}{3}$ and sum $1$.

The first term can be $\frac{32}{27}$ with common ratio $- \frac{1}{4}$ and sum $\frac{128}{135}$

These are the only two possibilities.

#### Explanation:

If the first term is $a$ and the common ratio is $r$, then we are given:

$a \left(1 + r\right) = \frac{8}{9}$

$a \left(1 + r + {r}^{2}\right) = \frac{26}{27}$

So:

$\frac{13}{12} = \frac{26 \cdot 9}{27 \cdot 8} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{a}}} \left(1 + r + {r}^{2}\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{a}}} \left(1 + r\right)} = \frac{1 + r + {r}^{2}}{1 + r}$

Cross multiply to get:

$13 \left(1 + r\right) = 12 \left(1 + r + {r}^{2}\right)$

Rearrange to get:

$0 = 12 {r}^{2} - r - 1 = \left(3 r - 1\right) \left(4 r + 1\right)$

So:

$r = \frac{1}{3} \text{ }$ or $\text{ } r = - \frac{1}{4}$

If $r = \frac{1}{3}$ then $a = \frac{8}{9 \left(1 + r\right)} = \frac{2}{3}$

The sum of the whole series is then $\frac{\frac{2}{3}}{1 - \frac{1}{3}} = 1$

If $r = - \frac{1}{4}$ then $a = \frac{8}{9 \left(1 + r\right)} = \frac{32}{27}$

The sum of the whole series is then $\frac{\frac{32}{27}}{1 - \left(- \frac{1}{4}\right)} = \frac{128}{135}$