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

`A_n = A_1 + d(n - 1)`

`A_1 = -32`

`A_9 = -120`

`=> A_9 = -32 + d(9 - 1)`

`=> -120 = -32 + d(8)`

`=> -88 = 8d`

`=> d = -11`

`A_32 = -32 + (-11)(32 - 1)`

`=> A_32 = -32 + (-11)(31)`

`=> A_32 = -32 + -341`

`=> A_32 = -373`

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

`color(blue)(a_n = a_1 + (n-1)d)`

So if can find the value of `a_1 and d` we can find any term,

In this case we are told that `a_1 = -32`

For the `9th` term, `n= 9` and `a_ = -120`

Use these values in the general formula.

`color(white)(......)color(blue)(a_ncolor(white)(...) = color(white)(...)a_1 color(white)(.......)+(n-1)d)`
`color(white)(......)uarrcolor(white)(..........)uarrcolor(white)(...............)uarrcolor(white)(...)uarr`
`" "-120color(white)(....)-32color(white)(..............)8color(white)(....)?`

This gives the equation:

`-120 = -32 +8d`

`-120+32 = 8d`

`-88 = 8d`

`d = -11`

Now the formula for the general term becomes:

`color(blue)(a_n = -32 + (n-1)(-11))" "larr` simplify

`color(blue)(a_n = -32 + -11n+11)`

`color(blue)(a_n = -11n -21)`

Find the `32nd` term (`n =32`)

`a_32 = -11(32)-21`

`a_32 = -352-21`

`a_32 = -373`