What is the 32nd term of the arithmetic sequence where a1 = 12 and a13 = –60?
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
To find the 32nd term of an arithmetic sequence, you need to know the first term (a1) and the common difference (d).
In this case, you are given that a1 = 12, but you need to determine the common difference (d).
To find the common difference, you can use the formula for the nth term of an arithmetic sequence:
an = a1 + (n-1)d
Using the given information, you can substitute the values:
a13 = a1 + (13-1)d
-60 = 12 + 12d
Simplify the equation:
-72 = 12d
Divide by 12 on both sides:
-6 = d
Now that you know the common difference (d = -6), you can use the formula for the nth term to find the 32nd term:
a32 = a1 + (32-1)d
a32 = 12 + 31(-6)
a32 = 12 - 186
a32 = -174
Therefore, the 32nd term of the arithmetic sequence is -174.