# What is the 9th term of the geometric sequence where a1 = -8 and a6 = -8,192?

May 31, 2016

We must first find the common ratio $r$.

#### Explanation:

The $n t h$ term of a geometric series is given by ${t}_{n} = a \times {r}^{n - 1}$. We know $a$, the first term, we know ${t}_{6}$, and therefore we know $n \left(6\right)$

$- 8192 = - 8 \times {r}^{6 - 1}$

$- \frac{8192}{-} 8 = {r}^{5}$

$1024 = {r}^{5}$

$\sqrt[5]{1024} = r$

$4 = r$

Now, we can reuse the formula ${t}_{n} = a \times {r}^{n - 1}$ to find the 9th term.

${t}_{9} = - 8 \times {4}^{9 - 1}$

${t}_{9} = - 8 \times {4}^{8}$

${t}_{9} = - 524288$

Therefore, ${t}_{9}$ is -524 288.

Practice exercises:

1. Determine the 11th term of a geometric sequence where the first term is 27 and ${t}_{4}$ is 1.

2. Determine the 7th term of a geometric sequence where the second term is 3 and the fifth is 375.

Hopefully this helps, and Good luck!