Let the amount of loan be P, and the rate of interest be
at r%, and the amount of repayment A in n years.
Then
after the 1st year P becomes (1+r/100)*P and we repay an amount
A.
So the next year this balance of (1+r/100)P-A) grows
with interest ((1+r/100)P-A)(1+r/100) and we repay A and the balance for for the 3rd
year = (1+r/100)^2*P - A(1+r/100) - A. In this way at the end of n th year the position
is :
(1+r/100)^n P - A{(1+r/100)^(n-1)+(1+r/100)(n-2)
+..+(1+r/100) + 1} which should be zero.
Or (1+r/100)^nP =
A{(1+r/100)^(n -1}/{1+r/100 -1}.
Solving for A , we
get:
A = (1+r/100)^n*p}(r/100)/ {1+r/100)^n
-1}...(1)
We know P= $100000,n = 10 years and r = 10%=
10/100. So, 1+10/100 = 1.1
Therefore, A =
{1.1^10*10^5}(0.1)/{1.1^10-1}
A = $16274.54 (approximated
to two decimals).
Therefore , in order to settle the loan
of $100000 in 10 years, we have to make the annual repayment of
$16274.54.
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