Kenton borrows $250,000 on January 1, 2012 to be repaid in 12 annual installments at an effective annual rate of interest of 12%. The first payment is due on January 1, 2013. Instead of annual payment he decides to make monthly payments equal to one-twelfth the annual payment beginning on February 1, 2013. Determine how many months will be needed to pay off the loan. (Ans : 129.381291)
Anyone know how to solve this question?
On
I misread the question and have revised
He makes the annual payment on 1/1/2013 and then starts paying monthly thereafter.
the annual payment:
$250,000 = P\sum_\limits{n=1}^{12} (1.12)^{-n}\\ 250,000 = P \frac {(1 - 1.12^{-12})}{0.12}\\ P = 250,000 \frac {0.01}{1-1.12^{-12}}\\ P = 40,359$
after making his first months payment he owes
$250,000 \frac {1-1.12^{-11}}{1-1.12^{-12}} = 239,640$
and he will be paying $\frac {40359}{12} = 3363$ per month
$239,640 = 3363\sum_\limits{n=1}^k (1.12)^{\frac {k}{12}}\\ 71.25 = \frac {1 - 1.12^{\frac {-k}{12}}}{(1.12)^{\frac 1{12}} - 1}\\ 1-71.25(1.12^\frac{1}{12} - 1) = (1.12)^{\frac {-k}{12}}\\ \ln 0.324 = \frac {-k}{12}\ln 1.12\\ 119.368 = k$
But $k$ doesn't start counting until one year has passed
$131$ months
Following that would be to use the annuity formula
$239641 = 3363*\dfrac{(1-\frac{1}{(1+r)^t})}{r}$
$\frac{1}{(1+r)^t} = 0.294671$ with r = 0.009489.
$(1+r)^t = 3.393617$
Taking log
$t \times ln(1.009489) = ln(3.393617)$
$t = \frac{ln(3.393617)}{ln(1.009489} = 129.3826$