# The third term of a geometric sequence is 3 and the sixth term is 64/9. How do you find the fifth term of this sequence?

##### 1 Answer
Mar 23, 2016

Write a systems of équations using the formula ${t}_{n} = a \times {r}^{n - 1}$

#### Explanation:

First equation:

$3 = a \times {r}^{3 - 1}$

Second equation:

$\frac{64}{9} = a \times {r}^{6 - 1}$

Solve by substitution:

$\frac{3}{r} ^ 2 = a \to \frac{64}{9} = \frac{3}{r} ^ 2 \times {r}^{5}$

$\frac{64}{9} = \frac{3 {r}^{5}}{r} ^ 2$

We can simplify further using the exponent rule ${a}^{x} / {a}^{m} = {a}^{x - m}$

$\frac{64}{9} = 3 {r}^{3}$

$\frac{\frac{64}{9}}{3} = {r}^{3}$

$\frac{64}{27} = {r}^{3}$

$\sqrt[3]{\frac{64}{27}} = r$

$\frac{4}{3} = r$

$3 = a \times {\left(\frac{4}{3}\right)}^{2}$

$3 = \frac{16}{9} a$

$3 \times \left(\frac{9}{16}\right) = a$

$\frac{27}{16} = a$

${t}_{n} = a \times {r}^{n - 1}$

${t}_{5} = \frac{27}{16} \times {\left(\frac{4}{3}\right)}^{4}$

${t}_{5} = \frac{27}{16} \times \frac{216}{81}$

${t}_{5} = \frac{9}{2}$

The 5th term is $\frac{9}{2}$