# What is the 32nd term of the arithmetic sequence where a1 = –32 and a9 = –120?

Nov 8, 2015

${A}_{32} = - 373$

#### Explanation:

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

${A}_{1} = - 32$

${A}_{9} = - 120$

$\implies {A}_{9} = - 32 + d \left(9 - 1\right)$

$\implies - 120 = - 32 + d \left(8\right)$

$\implies - 88 = 8 d$

$\implies d = - 11$

${A}_{32} = - 32 + \left(- 11\right) \left(32 - 1\right)$

$\implies {A}_{32} = - 32 + \left(- 11\right) \left(31\right)$

$\implies {A}_{32} = - 32 + - 341$

$\implies {A}_{32} = - 373$

Jul 29, 2017

${a}_{32} = - 373$

#### Explanation:

Any term in an arithmetic sequence can be found from the formula

$\textcolor{b l u e}{{a}_{n} = {a}_{1} + \left(n - 1\right) d}$

So if can find the value of ${a}_{1} \mathmr{and} d$ we can find any term,

In this case we are told that ${a}_{1} = - 32$

For the $9 t h$ term, $n = 9$ and ${a}_{=} - 120$

Use these values in the general formula.

$\textcolor{w h i t e}{\ldots \ldots} \textcolor{b l u e}{{a}_{n} \textcolor{w h i t e}{\ldots} = \textcolor{w h i t e}{\ldots} {a}_{1} \textcolor{w h i t e}{\ldots \ldots .} + \left(n - 1\right) d}$
$\textcolor{w h i t e}{\ldots \ldots} \uparrow \textcolor{w h i t e}{\ldots \ldots \ldots .} \uparrow \textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots} \uparrow \textcolor{w h i t e}{\ldots} \uparrow$
" "-120color(white)(....)-32color(white)(..............)8color(white)(....)?

This gives the equation:

$- 120 = - 32 + 8 d$

$- 120 + 32 = 8 d$

$- 88 = 8 d$

$d = - 11$

Now the formula for the general term becomes:

$\textcolor{b l u e}{{a}_{n} = - 32 + \left(n - 1\right) \left(- 11\right)} \text{ } \leftarrow$ simplify

$\textcolor{b l u e}{{a}_{n} = - 32 + - 11 n + 11}$

$\textcolor{b l u e}{{a}_{n} = - 11 n - 21}$

Find the $32 n d$ term ($n = 32$)

${a}_{32} = - 11 \left(32\right) - 21$

${a}_{32} = - 352 - 21$

${a}_{32} = - 373$