# What is the 8th term of the geometric sequence where a1 = 625 and a3 = 25?

Dec 20, 2015

It could be $\frac{1}{125}$ or $- \frac{1}{125}$

#### Explanation:

The general term of this sequence is of the form:

${a}_{n} = a {r}^{n - 1}$

where $a = 625$ is the initial term and $r$ the common ratio.

Then:

$25 = {a}_{3} = a {r}^{2} = 625 {r}^{2}$

So ${r}^{2} = \frac{25}{625} = \frac{1}{25}$

That results in two possibilities for $r$, viz $r = \frac{1}{5}$ and $r = - \frac{1}{5}$

If $r = \frac{1}{5}$ then:

${a}_{8} = 625 \cdot {\left(\frac{1}{5}\right)}^{7} = {5}^{4 - 7} = {5}^{- 3} = \frac{1}{125}$

If $r = - \frac{1}{5}$ then:

${a}_{8} = 625 \cdot {\left(- \frac{1}{5}\right)}^{7} = {\left(- 1\right)}^{7} {5}^{4 - 7} = - {5}^{-} 3 = - \frac{1}{125}$