What is the 32nd term of the arithmetic sequence where a1 = 12 and a13 = –60?

Answer 1

An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference of any two successive members of the sequence is a constant.

If the initial term of an arithmetic progression is a1 and the common difference of successive members is d, then the nth term of the sequence is given by:

an = a1 + ( n - 1 ) * d

In this case:

a13 = 12 + ( 13 - 1 ) * d

- 60 = 12 + 12 * d

- 60 - 12 = 12 d

- 72 = 12 d Divide both sides with 12

- 72 / 12 = 12 d / 12

- 6 = d

d = - 6

an = a1 + ( n - 1 ) * d

a32 = 12 + ( 32 - 1 ) * ( - 6 )

a32 = 12 + 31 * ( - 6)

a32 = 12 + ( - 186 )

a32 = 12 - 186

a32 = -174