Mr. Ade borrows ₦ 530.00 in the first month. Each month after that, he borrows ₦ 15.00 more than the previous month. How much will he have borrowed in total after 12 months?

The sequence of borrowings is ₦ 530.00, ₦ 545.00, ₦ 560.00, ...
The first term ($a$) is the amount borrowed in the first month:

$$a = 530$$

The common difference ($d$) is the difference between consecutive terms:

$$d = 545 - 530 = 15$$ $$d = 560 - 545 = 15$$

The total debt is for "a year", which means 12 months. So, the number of terms ($n$) is:

$$n = 12$$

The sum of the first $n$ terms of an arithmetic progression ($S_n$) is given by the formula:

$$S_n = \frac{n}{2}[2a + (n-1)d]$$

Substitute $a=530$, $d=15$, and $n=12$ into the formula:

$$S_{12} = \frac{12}{2}[2(530) + (12-1)15]$$ $$S_{12} = 6[1060 + (11)15]$$ $$S_{12} = 6[1060 + 165]$$ $$S_{12} = 6[1225]$$ $$S_{12} = \boxed{\text{₦ 7350.00}}$$