# How do you write a rule for the nth term of the arithmetic sequence a_11=29, a_20=101?

Aug 19, 2017

${a}_{n} = 8 n - 59$

#### Explanation:

Let set up two equation base on the information that given
${a}_{11} = 29 n = 11$ , $29 = {a}_{1} + \left(11 - 1\right) d$ , $29 = {a}_{1} + 10 d$

${a}_{20} = 101$, $101 = {a}_{1} + \left(20 - 1\right) d$ , $101 = {a}_{1} + 19 d$

Now we got:
$101 = {a}_{1} + 19 d$
$29 = {a}_{1} + 10 d$
We can solve this by using elimination method
So we need to multiply either equation to isolate ${a}_{1}$
$101 = {a}_{1} + 19 d$
$- 29 = - {a}_{1} - 10 d$
$72 = 9 d$ We can find d (common ratio) by divide 9 on both sides
$d = 8$
Now substitute d into either equation that we set up to find ${a}_{1}$
$29 = {a}_{1} + 10 \left(8\right)$
$29 = {a}_{1} + 80$
${a}_{1} = - 51$

The rule for an Arithmetic Sequence is ${a}_{n} = {a}_{1} + \left(n - 1\right) d$
$d = 8$ ${a}_{1} = - 51$
${a}_{n} = - 51 + \left(n - 1\right) 8$
${a}_{n} = - 51 + 8 n - 8$
${a}_{n} = 8 n - 59$