# How do you write a rule for the nth term of the arithmetic sequence a_8=19/5, a_12=27/5?

Mar 8, 2017

${a}_{n} = \frac{2}{5} n + \frac{3}{5}$

#### Explanation:

For the standard $\textcolor{b l u e}{\text{arithmetic sequence}}$

$a , a + d , a + 2 d , a + 3 d , \ldots \ldots . . ,$

The nth term is • a_n=a+(n-1)d

where a is the first term and d, the common difference.

For the sequence in question we have to find a and d

$\Rightarrow {a}_{8} = a + 7 d = \frac{19}{5} \to \left(\textcolor{red}{1}\right)$

$\Rightarrow {a}_{12} = a + 11 d = \frac{27}{5} \to \left(\textcolor{red}{2}\right)$

$\left(\textcolor{red}{2}\right) - \left(\textcolor{red}{1}\right)$ gives d

$\Rightarrow 4 d = \frac{8}{5} \Rightarrow d = \frac{2}{5}$

Substitute this value into $\left(\textcolor{red}{1}\right)$

$\Rightarrow a + \frac{14}{5} = \frac{19}{5} \Rightarrow a = 1$

We can now obtain the nth term rule.

$\Rightarrow {a}_{n} = 1 + \frac{2}{5} \left(n - 1\right)$

$\Rightarrow {a}_{n} = \frac{2}{5} n + \frac{3}{5} \leftarrow \textcolor{red}{\text{ nth term rule}}$