# How do you write a rule for the nth term of the arithmetic sequence a_10=2, a_22=-8.8?

Jul 14, 2018

${a}_{n} = 11 - 0.9 n$

#### Explanation:

The $n$-th term of an arihmetic progression is defined as:

${a}_{n} = {a}_{1} + \left(n - 1\right) d$

where $d$ is the common difference.

We know that

${a}_{10} = 2$
${a}_{22} = - 8.8$

$\therefore \left\{\begin{matrix}{a}_{1} + 9 d = 2 \\ {a}_{1} + 21 d = - 8.8\end{matrix}\right.$

To find the general rule, we have to solve this system of linear equation. It is fairly easy; substract the first equation from the second:

$\left({a}_{1} + 21 d\right) - \left({a}_{1} + 9 d\right) = - 10.8$

$12 d = - 10.8 \implies d = - 0.9$

Now substitute $d$ in any of the two equations to find ${a}_{1}$.

${a}_{1} - 8.1 = 2 \implies {a}_{1} = 10.1$

Hence the general rule of the $n$-term is going to be

${a}_{n} = 10.1 - 0.9 \left(n - 1\right)$

Which is equivalent to

${a}_{n} = 11 - 0.9 n$