# How do you write a rule for the nth term of the geometric sequence and then find a_5 given a_4=1/27, r=4/3?

Jul 30, 2017

${a}_{n} = {a}_{1} {r}^{n - 1}$
${a}_{5} = \frac{4}{81}$

#### Explanation:

The ${n}^{t h}$ term of a geometric sequence with first term ${a}_{1}$ and common ratio $r$ is given by:

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

In this example, ${a}_{4} = \frac{1}{27} \mathmr{and} r = \frac{4}{3}$

$\therefore \frac{1}{27} = {a}_{1} \times {\left(\frac{4}{3}\right)}^{3}$

$\frac{1}{27} = {a}_{1} \times {4}^{3} / {3}^{3}$

$\frac{1}{27} = {a}_{1} \times \frac{64}{27}$

$64 {a}_{1} = \frac{27}{27}$

${a}_{1} = \frac{1}{64}$

We are asked to find ${a}_{5}$ using the formula for the ${n}^{t h}$ term above.

${a}_{5} = {a}_{1} \times {r}^{4}$

$= \frac{1}{64} \times {\left(\frac{4}{3}\right)}^{4}$

$= \frac{1}{64} \times \frac{256}{81}$

$= \frac{4}{81}$

NB: We could have found this result more simply by using:

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

$\therefore {a}_{5} = {a}_{4} \times r$

${a}_{5} = \frac{1}{27} \times \frac{4}{3} = \frac{4}{81}$