Precalculus, Chapter 9, 9.2, Section 9.2, Problem 15

The given formula of nth term of a sequence is:
a_n = 3-4(n-2)
To solve for the first five terms of a sequence, plug-in the following values of n to the formula.
For the first term, plug-in n=1.
a_1=3-4(1-2) = 7
For the second term, plug-in n=2.
a_2=3-4(2-2)=3
For the third term, plug-in n=3.
a_3=3-4(3-2)=-1
For the second term, plug-in n=4.
a_4=3-4(4-2)=-5
And for the fifth term, plug-in n=5.
a_5=3-4(5-2)=-9
Therefore, the first five terms of the sequence are a_n={7, 3, -1, -5, -9,...}.
To determine if it is an arithmetic sequence, subtract a pair of terms. The pair should be consecutive terms. If there is a common difference, then it is an arithmetic sequence.
d= a_2-a_1 = 3-7=-4
d=a_3-a_2=-1-3=-4
d=a_4-a_3=-5-(-1)=-4
d=a_5-a_4=-9-(-5)=-4
Thus, the given a_n is an arithmetic sequence and its common difference between consecutive terms is -4.