# How do you write a rule for the nth term of the arithmetic sequence given a_7=6/7, a_9=2/3?

Feb 27, 2017

${a}_{n} = \frac{32 - 2 n}{21}$

#### Explanation:

${a}_{8}$ would be $\frac{1}{2} \left({a}_{7} + {a}_{9}\right)$

=$\frac{1}{2} \left(\frac{6}{7} + \frac{2}{3}\right) = \frac{16}{21}$

Common difference 'd' would thus be$\left(\frac{16}{21} - \frac{6}{7}\right) = - \frac{2}{21}$

If the first term is say 'x, then 7th term ${a}_{7} = x + \left(7 - 1\right) \left(- \frac{2}{21}\right)$

Thus $\frac{6}{7} = x - \frac{4}{7}$, Or $x = \frac{10}{7}$

nth term expresssion ${a}_{n} = {a}_{1} + \left(n - 1\right) d$ can now be written as ${a}_{n} = \frac{10}{7} + \left(n - 1\right) \left(- \frac{2}{21}\right)$

If required, this can be put in a compact form as

${a}_{n} = \frac{- 2 n}{21} + \frac{32}{21} = \frac{32 - 2 n}{21}$