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Financial sequences and geometric sums

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    Name
    Vu Hung
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Problem Statement

A loan of \Listakenoutataninterestrateofis taken out at an interest rate ofrperperiod.Attheendofeachperiod,aconstantrepaymentofper period. At the end of each period, a constant repayment of$Pismadeafterinteresthasbeenaddedtothebalance.Letis made after interest has been added to the balance. LetM_nbethebalanceowingafterbe the balance owing aftern$ periods.

  • Write down expressions for M1M_1 and M2M_2 in terms of L,r,L,r, and PP.
  • Show that the balance after nn periods can be written as
Mn=L(1+r)n−P[1+(1+r)+(1+r)2+⋯+(1+r)n−1].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
Mn=L(1+r)n−P((1+r)n−1)r.M_n=L(1+r)^n-\frac{P((1+r)^n-1)}{r}.
  • A car loan of \20,000istakenatis taken at1%interestpermonth.Calculatethemonthlyrepaymentinterest per month. Calculate the monthly repaymentPrequiredtoreducethebalancetozeroattheendofrequired to reduce the balance to zero at the end of24$ months.

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 11 and ratio 1+r1+r.
  • For (iv): Set M24=0M_{24}=0, with L=20000L=20000 and r=0.01r=0.01.

Solutions

(i) After one period,

M1=L(1+r)−P.M_1=L(1+r)-P.

After two periods,

M2=(L(1+r)−P)(1+r)−P=L(1+r)2−P(1+r)−P.M_2=(L(1+r)-P)(1+r)-P=L(1+r)^2-P(1+r)-P.

(ii) Continuing this pattern gives

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

Reordering the bracket,

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

(iii) The bracket is a finite GP with first term 11, ratio 1+r1+r, and nn terms:

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

Hence

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

(iv) Here L=20000L=20000, r=0.01r=0.01, and n=24n=24. Set M24=0M_{24}=0:

0=20000(1.01)24−P((1.01)24−1)0.01.0=20000(1.01)^{24}-\frac{P((1.01)^{24}-1)}{0.01}.

Thus

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

Using (1.01)24≈1.269735(1.01)^{24}\approx1.269735,

P≈$941.47.P\approx \$941.47.

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.

Further Readings

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