# How do you write a rule for the nth term of the arithmetic sequence and then find a_22 given d=-11, a_13=99?

May 30, 2017

${a}_{n} = 231 - 11 \left(n - 1\right)$

${a}_{22} = 0$

#### Explanation:

First, find ${a}_{1}$:

${a}_{1} = {a}_{13} - 12 d$

${a}_{1} = 99 - 12 \left(- 11\right)$

${a}_{1} = 99 + 132 = 231$

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

The rule for the nth term is ${a}_{n} = 231 - 11 \left(n - 1\right)$

Finally, plug in $n = 22$ and solve.

${a}_{n} = 231 - 11 \left(n - 1\right)$

${a}_{22} = 231 - 11 \left(22 - 1\right)$

${a}_{22} = 231 - 11 \left(21\right)$

${a}_{22} = 231 - 231$

${a}_{22} = 0$