Question

What is the 32nd term of the arithmetic sequence where a1 = -33 and a9 = -121?

Answer 1

`a_32=-374`

Explanation:

An arithmetic sequence is of the form:
`a_(i+1)=a_i+q`

Therefore, we can also say:
`a_(i+2)=a_(i+1)+q=a_i+q+q=a_i+2q`

Thus, we can conclude:
`a_(i+n)=a_i+nq`

Here, we have:
`a_1=-33`
`a_9=-121 rarr a_(1+8)=-33+8q=-121`

`rarr 8q=-121+33=-88 rarr q=(-88)/8=-11`

Therefore:
`a_32=a_(1+31)=-33-11*31=-33-341=-374`