**What is the payment on a 12‐year, $400,000 mortgage that compounds on the payment date every 73 days at 6%? (show your formula and calculations, and ignore leap year)**

I do not really understand the question so here is my best guess. I am assuming I do not have to worry about individual days in a month. Also, I am assuming 6% is the annual rate.

**Step 1**: Calculate relationship between annual rate, 73 days rate and monthly rate.

365 Days = 12 months => 73 Days = 12*(73/365) months = 2.4 months

Let R be the annual rate in decimal (0.06 in this case); r be the monthly rate in decimal and R* is the rate every 73 days (or every 2.4 months)

To convert rates lets start with something familiar.

If a loan compounds every 6 months (half yearly) then 6 months rate is R/2 i.e. R/(12/6)

If a loan compounds every 3 months (quarterly) the 3 months rate is R/4 i.e. R/(12/3)

If a loan compounds every month then monthly rate is R / 12 i.e. R/(12/1)

So, if a loan compounds ever 2.4 months (73 days) the 73 day rate is R/(12/2.4) = R/5 = R*

Now, monthly rate r compounded for 2.4 months = R* i.e. (1+r)

^{2.4 }= 1 + R*

r = (1+R*)

^{(1/2.4)}– 1 = (1+(R/5))

^{(1/2.4)}– 1 = (1+(0.06/5))

^{(1/2.4)}– 1 = 0.004983 = 0.4983%

If the loan had compounded every month instead of 73 days then r = R/12 = 0.5%

**Step 2**: Derive Payment Formula

Let L be the loan amount ($400,000) and P be the payment every month

At the end of the 1st month:

Outstanding Loan Amount, L

_{1}= L*(1+r) – P

Let (1+r) = X

L

_{1}= L*X – P

At the end of 2nd month:

Outstanding Loan Amount, L

_{2}= L

_{1}*X – P = L*X

^{2}– P*X – P

At the end of 3rd month:

Outstanding Loan Amount, L

_{3}= L

_{2}*X – P = L*X

^{3}– P*(X

^{2}– X – 1)

At the end of nth month:

Outstanding Loan Amount, L

_{n}= L

_{n-1}*X – P = L*X

^{n}– P*(X

^{n-1}– X

^{n-2}– … 1)

Sum of Geometric Progression (GP) 1+x+x

^{2}+x

^{3}+…x

^{n-1}= (x

^{n}– 1)/(x – 1)

Using the GP formula and replacing X by (1+r)

L

_{n}= L*(1+r)

^{n}– P*(((1+r)

^{n}– 1) / r)

At the end of loan term (n = 144 months in this case) outstanding amount = 0

When L

_{n}= 0 then P = L*r*(1+r)

^{n}/ ((1+r)

^{n}– 1) = L*r / (1 – 1/(1+r)

^{n})

Therefore, P = 400,000*0.004983 / (1-1/(1+0.004983)

^{144}) = 3899.39

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