# How do you write a rule for the nth term of the geometric sequence given the two terms a_4=351, a_7=13?

May 17, 2017

${a}_{n} = 9477 \cdot {\left(\frac{1}{3}\right)}^{n - 1}$

#### Explanation:

We have: ${a}_{4} = 351$ and ${a}_{7} = 13$

The $n$th term of a geometric sequence is given by:

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

Let's express the $4$th and $7$th terms using this rule:

$R i g h t a r r o w {a}_{4} = 351$

$R i g h t a r r o w {a}_{1} {r}^{4 - 1} = 351$

$R i g h t a r r o w {a}_{1} {r}^{3} = 351$ ----------- $\left(i\right)$

and

$R i g h t a r r o w {a}_{7} = 13$

$R i g h t a r r o w {a}_{1} {r}^{7 - 1} = 13$

$R i g h t a r r o w {a}_{1} {r}^{6} = 13$ ------------ $\left(i i\right)$

Then, let's divide $\left(i i\right)$ by $\left(i\right)$:

$R i g h t a r r o w \frac{{a}_{1} {r}^{6}}{{a}_{1} {r}^{3}} = \frac{13}{351}$

$R i g h t a r r o w {r}^{3} = \frac{1}{27}$

$R i g h t a r r o w r = \frac{1}{3}$

Now, let's find the first term by substituting this value for the common ratio into $\left(i i\right)$:

$R i g h t a r r o w {a}_{1} {\left(\frac{1}{3}\right)}^{6} = 13$

$R i g h t a r r o w \frac{{a}_{1}}{729} = 13$

$R i g h t a r r o w {a}_{1} = 9477$

Finally, let's substitute these values back into the rule for the $n$th term:

$\therefore {a}_{n} = 9477 \cdot {\left(\frac{1}{3}\right)}^{n - 1}$