A house worth $\$180,000$ was purchased using a $\$30,000$ deposit under a loan agreement that stipulated monthly repayments of $\$1500$ and interest of 4% per annum on the unpaid balance. How much was eventually paid for the house?
My solution
Using arithmetic sequence sum:
loan = $\$180,000-\$30,000 = \$150,000$
rate per month r = 0.04/12 = 1/300
A=P(1+(r/100)^n)
A1=150000(301/300)-1500
A2=A1(301/300)-1500 = 150000(301/300)^2-1500(301/300)-1500
An = 150000 (301/300)^n - 1500 [1+301/300+.......+ 301/300^(n-1)]
An = 150000 (301/300)^n - 450000 [(301/300)^n -1]
let An = 0
n = log (3/2)/ log (301/300) = 121.84=122month
total payment = 1500 x 122month= 183000
total paid to the house = 183000 + 30000 = 213000
BUT answer = $\$205250$
You have not stated how you got $n$ or $d$, so it is not clear how you got $205250$. One way to solve this problem is in a spreadsheet. Make a column for month numbers, one for the balance at the start of the month, one for interest, one for payment. The interest is the starting balance times $\frac {0.04}{12}$. The payment is what it is. The balance at the start of the next month is the balance at the start of this month plus interest minus payment. Copy down and see how many months it takes to get the balance to zero. You might want to make the last month's payment be just what it takes to get to zero. Then add up the payments and you are done.
I suggest you put the interest rate in a fixed cell and use absolute reference to it. This spreadsheet will come in handy many times if you make it a bit flexible.
There should also be a formula in your book for this. It will be one of the types of annuity. You will have to read carefully to get the right formula.