# How do you write a rule for the nth term of the arithmetic sequence a_6=-31, a_14=-135?

Jun 7, 2017

${T}_{n} = - 13 n + 47$

#### Explanation:

For any term in an AP, the rule is the same: ${T}_{n} = a + \left(n - 1\right) d$

For the $6 t h$ term we have..
${T}_{6} = a + 5 d = - 31 \text{ } \leftarrow \left(6 - 1 = 5\right)$

For the $14 t h$ term we have:
${T}_{14} = a + 13 d = - 135 \text{ } \leftarrow \left(14 - 1 = 13\right)$

To be able to write the rule we need to find $a$ and $d$.
Solve the equations simultaneously:

$\textcolor{w h i t e}{\times \times \times} {T}_{6} = a + \text{ } 5 d = - 31$.........................................A
$\textcolor{w h i t e}{\times \times \times} {T}_{14} = a + \text{ } 13 d = - 135$.........................................B

A-B:$\textcolor{w h i t e}{w w w w w w w w w w} - 8 d = 104$
A-B:$\textcolor{w h i t e}{w w w w w w . w w w w w w} d = - 13$

Subst in A:$\rightarrow a + 5 \left(- 13\right) = - 31$
$\text{ "a" } - 65 = - 31$
$\text{ "a " } = - 31 + 65$
$\text{ "a " } = 34$

Now we know that the first term is $34$ and that $d = - 13$

The first $6$ terms are: $34 , \text{ "21," } 8 , - 5 , - 18 , - 31 \ldots . .$

The rule for the $n t h$ term is:

${T}_{n} = 34 + \left(n - 1\right) \left(- 13\right)$

${T}_{n} = 34 - 13 n + 13$

${T}_{n} = - 13 n + 47$