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

1 Answer
Jul 4, 2015

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