· H2 Mathematics — Ten-Year Series, worked ← All 11 questions

2023 A-Level · Paper 1 · Question 11

2023 H2 Maths Paper 1, Question 11

Sequences and Series 13 marks Arithmetic/Geometric Progression
Download Paper 1 solutions (PDF)

Part (a)

2 marks

Approach: the deposits form an arithmetic progression; because the condition is “at least $50 000\$ 50\, 000” and aa is wanted to the nearest cent, we solve the equality first and then check whether rounding up or down is what actually satisfies the inequality.

From 31 January 2021 to 31 December 2023 is exactly 3 years, i.e. 3636 monthly deposits, forming an arithmetic progression with first term $a\$ a, common difference $50\$ 50, and 3636 terms.

S36=362[2a+(36−1)(50)]S_{36} = \frac{36}{2}\left\lbrack 2a + (36 - 1)(50) \right\rbrack
=18(2a+1750)= 18(2a + 1750)
=36a+31500= 36a + 31500

Requiring S36≥50 000S_{36} \geq 50\, 000,

36a+31500≥50 00036a + 31500 \geq 50\, 000
⇒36a≥18 500\Rightarrow 36a \geq 18\, 500
⇒a≥18 50036\Rightarrow a \geq \frac{18\, 500}{36}
=513.88‾= 513.8\overline{8}

Since aa is to the nearest cent, we must round up (rounding down would leave S36<50 000S_{36} < 50\, 000).

To check your answer: at a=$513.89a = \$ 513.89, S36=36(513.89)+31500=$50 000.04≥50 000S_{36} = 36(513.89) + 31500 = \$ 50\, 000.04 \geq 50\, 000; at a=$513.88a = \$ 513.88, S36=$49 999.68<50 000S_{36} = \$ 49\, 999.68 < 50\, 000 — confirming $513.89\$ 513.89 is indeed the smallest cent value that works. ✓

∴\therefore the smallest value of aa is $513.89\$ 513.89.

Part (b)

3 marks

Approach: set up a recurrence for the amount owed, unroll it for the first few months to see the pattern, then sum the resulting geometric series in general.

Let LnL_{n} be the amount owed after nn months. Each month, interest of 0.1%=0.0010.1\% = 0.001 is added first, then the repayment xx is subtracted:

Ln=1.001Ln−1−x,L0=400 000L_{n} = 1.001L_{n - 1} - x,\quad L_{0} = 400\, 000

Applying this repeatedly,

L1=1.001L0−xL_{1} = 1.001L_{0} - x
L2=1.001L1−xL_{2} = 1.001L_{1} - x
=(1.001)2L0−x(1.001+1)= (1.001)^{2}L_{0} - x(1.001 + 1)
L3=1.001L2−xL_{3} = 1.001L_{2} - x
=(1.001)3L0−x((1.001)2+1.001+1)= (1.001)^{3}L_{0} - x\left( (1.001)^{2} + 1.001 + 1 \right)

In general,

Ln=(1.001)nL0−x((1.001)n−1+(1.001)n−2+⋯+1.001+1)L_{n} = (1.001)^{n}L_{0} - x\left( (1.001)^{n - 1} + (1.001)^{n - 2} + \cdots + 1.001 + 1 \right)

The bracketed term is a geometric series of nn terms with first term 11 and common ratio 1.0011.001:

(1.001)n−1+⋯+1(1.001)^{n - 1} + \cdots + 1
=(1.001)n−11.001−1= \frac{(1.001)^{n} - 1}{1.001 - 1}
=1000((1.001)n−1)= 1000\left( (1.001)^{n} - 1 \right)

Substituting back,

Ln=400 000(1.001)n−1000x((1.001)n−1)L_{n} = 400\, 000(1.001)^{n} - 1000x\left( (1.001)^{n} - 1 \right)
To check your answer: at n=1n = 1, the formula gives 400 000(1.001)−1000x(1.001−1)=400 400−x400\, 000(1.001) - 1000x(1.001 - 1) = 400\, 400 - x, matching the recurrence L1=1.001(400 000)−x=400 400−xL_{1} = 1.001(400\, 000) - x = 400\, 400 - x directly. ✓

as required.  (shown)

Part (c)(i)

4 marks

For the loan to be fully repaid in 360360 months, L360=0L_{360} = 0:

400 000(1.001)360−1000x((1.001)360−1)=0400\, 000(1.001)^{360} - 1000x\left( (1.001)^{360} - 1 \right) = 0
⇒1000x((1.001)360−1)=400 000(1.001)360\Rightarrow 1000x\left( (1.001)^{360} - 1 \right) = 400\, 000(1.001)^{360}
x=400(1.001)360(1.001)360−1x = \frac{400(1.001)^{360}}{(1.001)^{360} - 1}
≈1323.6348\approx 1323.6348
≈$1323.63\approx \$ 1323.63

∴\therefore the monthly repayment is $1323.63\$ 1323.63 (nearest cent).

Part (c)(ii)

Using the rounded repayment from (c)(i), the total repaid over 360360 months is 360×$1323.63360 \times \$ 1323.63. The total interest paid is this total minus the original loan amount:

Total interest=360(1323.63)−400 000\text{Total interest} = 360(1323.63) - 400\, 000
=476 506.80−400 000= 476\, 506.80 - 400\, 000
=$76 506.80= \$ 76\, 506.80

∴\therefore the total interest paid is $76 506.80\$ 76\, 506.80.

Part (d)(i)

4 marks

Approach: with a fixed $1600\$ 1600 payment, use the formula from (b) to write LnL_{n} explicitly, find the (non-integer) nn at which Ln=0L_{n} = 0, and note that Wei cannot overpay — so she pays $1600\$ 1600 for the largest whole number of months kk that still leaves a positive balance, then clears the rest with a smaller final payment yy.

With x=1600x = 1600 in the formula from (b),

Ln=400 000(1.001)n−1 600 000((1.001)n−1)L_{n} = 400\, 000(1.001)^{n} - 1\, 600\, 000\left( (1.001)^{n} - 1 \right)
=1 600 000−1 200 000(1.001)n= 1\, 600\, 000 - 1\, 200\, 000(1.001)^{n}

LnL_{n} is decreasing (since (1.001)n(1.001)^{n} grows), reaching 00 when

1 600 000−1 200 000(1.001)n=01\, 600\, 000 - 1\, 200\, 000(1.001)^{n} = 0
⇒(1.001)n=43\Rightarrow (1.001)^{n} = \frac{4}{3}
⇒n=ln⁡(4/3)ln⁡(1.001)\Rightarrow n = \frac{\ln(4/3)}{\ln(1.001)}
≈287.83\approx 287.83

Since nn is not a whole number, Wei makes k=287k = 287 full payments of $1600\$ 1600, leaving a small positive balance that is cleared by a smaller final payment.

Note: a 288288th payment of $1600\$ 1600 would overpay, since 287.83<288287.83 < 288.

L287=1 600 000−1 200 000(1.001)287L_{287} = 1\, 600\, 000 - 1\, 200\, 000(1.001)^{287}
≈1320.22\approx 1320.22

The final repayment clears this balance after one more month’s interest is added:

y=1.001 L287y = 1.001\, L_{287}
≈1321.54\approx 1321.54
To check your answer: y≈1321.54<1600y \approx 1321.54 < 1600 as required; a further $1600\$ 1600 payment at month 288288 instead would give 1.001(1320.22)−1600≈−278.461.001(1320.22) - 1600 \approx - 278.46, i.e. an overpayment, confirming k=287k = 287 and not 288288. ✓

∴\therefore k=287k = 287 and y≈$1321.54y \approx \$ 1321.54.

Part (d)(ii)

The total repaid under the plan in (d)(i):

1600(287)+1321.54=460 521.541600(287) + 1321.54 = 460\, 521.54

The total repaid under the plan in (c)(i):

360(1323.63)=476 506.80360(1323.63) = 476\, 506.80
Saving=476 506.80−460 521.54\text{Saving} = 476\, 506.80 - 460\, 521.54
≈$15 985.26\approx \$ 15\, 985.26
≈$15 990 (4 s.f.)\approx \$ 15\, 990\ \left( 4\text{ s.f.} \right)

∴\therefore Wei saves approximately $15 990\$ 15\, 990 (4 s.f.) by making the repayments in (d)(i) rather than in (c)(i).

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