A reducing loan and a final smaller payment · source reference

Printed question

Selena borrows 200000 dollars from a bank for a home purchase on 1 January 2025. On day 1 of each month, 0.5% interest is added, first on 1 January 2025. On each last day she pays $x$ dollars, starting 31 January 2025.

(i) Show debt at end of $n$ months is $200000(1.005)^n-200x(1.005^n-1)$. [3]

(ii) If she pays 1500 dollars monthly on each last day, on what date does she fully repay, and what is the last repayment amount? [4]

(iii) To fully repay in 10 years, how much monthly? [2]

Printed working

(i) End of $n$ months:

$$\begin{aligned}B_n&=200000(1.005)^n-x[1+1.005+(1.005)^2+\cdots+(1.005)^{n-1}]\\&=200000(1.005)^n-x\frac{1-1.005^n}{1-1.005}\\&=200000(1.005)^n-200x[(1.005)^n-1].\end{aligned}$$

(ii) At $x=1500$,

$$200000(1.005)^n-200(1500)[(1.005)^n-1]\le0\Rightarrow300000-100000(1.005)^n\le0,$$ $$(1.005)^n\ge3\Rightarrow n\ge\frac{\ln3}{\ln1.005}\ge220.27.$$

Least $n=221$; alternatively GC table below.

At end of 220 months owed $200000(1.005)^{220}-200(1500)[(1.005)^{220}-1]=405.67$ dollars. Last payment in month 221 $=405.67(1.005)=407.70$ (2 d.p.). 221 months = 18 years 5 months: fully paid on 31 May 2043.

(iii) Ten years =120 months:

$$200000(1.005)^{120}-200x[(1.005)^{120}-1]\le0,$$ $$363879.3468-163.8793468x\le0\Rightarrow x\ge2220.410039.$$

Pay 2220.42 dollars each month (2 d.p.); 2220.41 not accepted.

Source tables and calculator displays

Month	Amount at beginning	Amount at end
1	$200000(1.005)$	$200000(1.005)-x$
2	$200000(1.005)^2-(1.005)x$	$200000(1.005)^2-(1.005)x-x$
3	$200000(1.005)^3-(1.005)^2x-(1.005)x$	$200000(1.005)^3-(1.005)^2x-(1.005)x-x$

$n$	Debt after payment
219	1896.19
220	405.67 >0
221	−1092.30 <0

Printed answer key

(ii) 31 May 2043; 407.70 dollars (2 d.p.); (iii) 2220.42 dollars (2 d.p.)

Mathematical review and qualifications

(i) With $B_0=200000$, unrolling gives $B_n=200000q^n-x\sum_{j=0}^{n-1}q^j=200000q^n-200x(q^n-1)$, $q=1.005$.

(ii) At $x=1500$, $B_n=300000-100000q^n$. The condition $B_n\le0$ gives $n\ge\ln3/\ln1.005\approx220.271$, so month 221. After 220 payments debt is about 405.67234; after the next interest charge the final amount is $qB_{220}\approx407.70070$, or 407.70 dollars. Month 221 ends on 31 May 2043.

(iii) For 120 months, $x\ge1000q^{120}/(q^{120}-1)\approx2220.410039$. To guarantee clearance with payments in whole cents, round upward to 2220.42 dollars. The ordinary nearest-cent value 2220.41 falls slightly short.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. [2024/JPJC/Prelim/I/8] Question PDF pages 6; solution PDF pages 26, 27. The original files remain private QA records outside this native content.