Answer :
To solve this problem, we need to utilize the concept of amortization schedules where loans are paid off through fixed monthly payments. These payments consist of both principal and interest components.
Given:
- Annual nominal rate = 9% compounded monthly
- Amount of principal repaid in the 12th payment = 1000
- Amount of principal repaid in the Tth payment = 3700
The key to solving the problem is recognizing that each monthly payment is fixed, but the amount dedicated to interest and principal changes over time. At the beginning of a loan, payments largely cover interest; toward the end, they pay down more principal.
Let's represent the monthly loan interest rate as [tex]i[/tex]:
[tex]i = \frac{0.09}{12} = 0.0075.[/tex]
To further analyze this, we can use the formula for the principal portion of an installment in an amortized loan, often expressed in the scenario of such problems as:
[tex]P_k = C(1+i)^{1-k}[/tex]
where:
- [tex]P_k[/tex] is the principal repaid in the k-th payment.
- [tex]C[/tex] is a constant that represents the loan amount multiplier in amortized form.
From the problem, we know:
[tex]P_{12} = C(1.0075)^{1-12} = 1000[/tex]
[tex]P_{T} = C(1.0075)^{1-T} = 3700[/tex]
To find [tex]T[/tex], divide these two equations:
[tex]\frac{C(1.0075)^{1-12}}{C(1.0075)^{1-T}} = \frac{1000}{3700}[/tex]
Simplifying gives:
[tex](1.0075)^{T-12} = 3.7[/tex]
Taking the natural logarithm of both sides:
[tex]\ln((1.0075)^{T-12}) = \ln(3.7)[/tex]
[tex](T-12)\ln(1.0075) = \ln(3.7)[/tex]
Solving for [tex]T[/tex]:
[tex]T-12 = \frac{\ln(3.7)}{\ln(1.0075)}[/tex]
Calculate the right side using logarithms:
[tex]T-12 \approx \frac{1.30833}{0.00748} \approx 174.91[/tex]
So, [tex]T[/tex] is approximately:
[tex]T = 174.91 + 12 \approx 186.91[/tex]
Rounding to the nearest whole number, [tex]T = 187[/tex].
Hence, the correct answer is option (b) [tex]187[/tex].
