10 weeks.
This is a simple problem which we can represent as a series as it happens over time with the same constant value of $1.5 per week.
The first term of the A.P. is 1.5
If the common difference is 1.5, find the term for which the term is 15$
nth term of an AP =
[tex]a \: + (n \: - 1)d[/tex]
d = 1.5,
a = 1.5
so if
15 = 1.5 + (n-1)1.5
we have that
15 = 1.5 + 1.5n - 1.5
1.5 - 1.5 = 0
So, 15 = 1.5n
Divide both sides by 1.5 to get the unity value of n
n =
[tex] \frac{15}{1.5} [/tex]
n = 10
