# How do you write a rule for the nth term of the geometric sequence, then find a6 given 2, 4/3, 8/9, 16/27,...?

May 31, 2016

${T}_{6} = \frac{64}{243}$

#### Explanation:

The first term is given as 2, so $a = 2$

The common ration, $r$ is found by dividing 2 consecutive terms..
$r = \frac{{T}_{n}}{{T}_{n - 1}}$ In other words, a term divided by the one before it.

r = 4/3 ÷2/1 = 4/3 xx 1/2 = 2/3

Check: r = 8/9 ÷ 4/3 = 8/9 xx 3/4 =2/3

Now, knowing the values for $a \mathmr{and} r$, we can substitute into the general form for a term of a GP

${T}_{n} = a {r}^{n - 1} \Rightarrow 2 \times {\left(\frac{2}{3}\right)}^{n - 1}$

Simplifying gives:  T_n =(2^n)/(3^(n-1)

In the same way we can find ${a}_{6} = a {r}^{5}$

But we have already found the general form, so let's use that for $n = 6$

${T}_{6} = \frac{{2}^{6}}{{3}^{5}}$

${T}_{6} = \frac{64}{243}$