Financial sequences and geometric sums

This problem explores financial sequences using geometric progressions, a key topic in the HSC Mathematics Extension 1 syllabus under Money matters. By modelling a reducing-balance loan with regular repayments, you’ll learn to connect recursive sequences to closed-form formulas and use the sum of a finite geometric series to solve for unknown repayments—an essential skill for tackling annuity and loan problems.

Problem Statement

A loan of $L$ is taken out at an interest rate of $r$ per period. At the end of each period, a constant repayment of $P$ is made after interest has been added to the balance. Let $M_n$ be the balance owing after $n$ periods.

• Write down expressions for $M_1$ and $M_2$ in terms of $L,r,$ and $P$.
• Show that the balance after $n$ periods can be written as

$$M_n=L(1+r)^n-P\left[1+(1+r)+(1+r)^2+\cdots+(1+r)^{n-1}\right].$$

• Using the formula for the sum of a geometric progression, prove that

$$M_n=L(1+r)^n-\frac{P((1+r)^n-1)}{r}.$$

• A car loan of $$20\,000$ is taken at $1\%$ interest per month. Calculate the monthly repayment $P$ required to reduce the balance to zero at the end of $24$ months.

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Hints

• For (i): Add interest first, then subtract the repayment.
• For (ii): Expand the first few balances and look for the pattern.
• For (iii): The bracket is a finite GP with first term $1$ and ratio $1+r$.
• For (iv): Set $M_{24}=0$, with $L=20000$ and $r=0.01$.

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Solutions

(i) After one period, the interest is applied to the initial loan amount, then the repayment is subtracted:

$$M_1=L(1+r)-P.$$

For the second period, we take the new balance $M_1$, apply interest again, and subtract the constant repayment $P$:

$$M_2=(L(1+r)-P)(1+r)-P=L(1+r)^2-P(1+r)-P.$$

(ii) The pattern from parts (i) reveals that each additional period introduces another factor of $(1+r)$ and subtracts $P$ times the accumulated sum of powers. Extending this recurrence $n$ times gives

$$M_n=L(1+r)^n-P\left[(1+r)^{n-1}+(1+r)^{n-2}+\cdots+(1+r)+1\right].$$

Reordering the terms inside the brackets to start from $1$,

$$M_n=L(1+r)^n-P\left[1+(1+r)+(1+r)^2+\cdots+(1+r)^{n-1}\right].$$

(iii) The expression in brackets is a finite geometric progression. Recall that the sum of a geometric series with $n$ terms, first term $a$, and common ratio $R$ is $\frac{a(R^n-1)}{R-1}$. Here $a=1$ and $R=1+r$, so we obtain

$$1+(1+r)+\cdots+(1+r)^{n-1} =\frac{(1+r)^n-1}{(1+r)-1} =\frac{(1+r)^n-1}{r}.$$

Hence

$$M_n=L(1+r)^n-\frac{P((1+r)^n-1)}{r}.$$

(iv) To clear the loan exactly after $24$ months, we set the final balance to zero and solve for the monthly repayment. Substituting $L=20000$, $r=0.01$, and $n=24$ into the formula gives

$$0=20000(1.01)^{24}-\frac{P((1.01)^{24}-1)}{0.01}.$$

Thus

$$P=\frac{20000(1.01)^{24}(0.01)}{(1.01)^{24}-1}.$$

Using $(1.01)^{24}\approx1.269735$,

$$P\approx \$941.47.$$

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Takeaways

• Loan balances can be modelled by applying interest and repayment recursively.
• Repeated repayments form a finite geometric sum after compounding.
• Setting the final balance to zero gives the repayment required to clear a loan.

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Further Readings

HSC Last Resorts, HSC Combinatorics, HSC Sequences, HSC Integrals