We are given the equation:
an= 2 - 5(n +1) --->
eqtn 1
The domain is defined to be all the values of the input
variable (in this case n) in which the function is defined. Or in other words, all
the values of n in which we can have reasonable value of an.
But before we find for the domain, let us first define
the variables. A standard arithmetic sequence or series has a general form of:
an = a1 + d (n – 1) --->
eqtn 2
where,
an = is the nth value specified by the value of n
a1 = is the 1st value or 1st term
d = is the common difference
n = the order of value
We can see from equation 2 that our a1 corresponds to an
n value of n = 1 so that the factor d (n – 1) is cancelled leaving us with a1
alone. So the domain starts from 1 to positive infinity or in other words, all
positive integers.
However in our equation 1, this is not the case, we can
only get a1 when n = -1, therefore our domain starts from -1. So the domain is:
-1 , 0 and all positive integers
(n ≥ -1)
