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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