Free worked solutions · 2026 Edition

IB Maths AI HL Chapter 2: Sequences, series and financial mathematics worked solutions

Worked solutions to the practice questions in Chapter 2 of Photon Guidebook: Mathematics Applications and Interpretation HL for the IB Diploma (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[3 marks]
Hana organises deliveries. The first 11 weekly deliveries contain 748 boxes in total. The first delivery contains 23 boxes and each week adds a constant number.
Find this weekly increase.
[3]
Show solutionHide solution
748=11[46+10d]/2
M1
136=46+10d
M1
d=9 boxes per week
A1

Final answer

d=9 boxes per week.

Question 2

[4 marks]
Ari measures markers. The heights of markers are h_n=18+2.4(n-1) cm, for positive integer n.
Find the first marker above 60 cm, and give its height.
[4]
Show solutionHide solution
18+2.4(n-1)>60 gives n>18.5.
M1
First integer n=19.
A1
h_{19}=18+18(2.4)
M1
h_{19}=61.2 cm (while h_{18}=58.8 cm).
A1

Final answer

Marker 19; 61.2 cm.

Question 3

[4 marks]
Nuri models bead counts in a geometric sequence that has u_3=28 and u_5=63, with positive ratio.
Find the common ratio and the fourth term.
[4]
Show solutionHide solution
r^2=63/28=2.25.
M1
r=1.5, selecting the positive root.
A1
u_4=28(1.5).
M1
u_4=42.
A1

Final answer

r=1.5; u_4=42.

Question 4

[3 marks]
Zoe tests a coating that begins with 340 units before treatment. Each treatment leaves 78% of the previous amount.
Find the smallest number of treatments leaving less than 60 units. Use neighbouring amounts.
[3]
Show solutionHide solution
Use 340(0.78)^n for integer n.
M1
At n=6: 76.6; at n=7: 59.7.
A1
7 treatments; the amounts decrease strictly.
A1

Final answer

7 treatments; 59.7 units.

Question 5

[3 marks]
Ben’s print press falls in value from 15 000 US dollars to 10 935 US dollars over three years under a constant annual percentage depreciation model.
Use a GDC equation solver to find the annual depreciation percentage.
[3]
Show solutionHide solution
Solve 15\,000(1-d/100)^3=10\,935 for 0<d<100.
M1
Retained yearly multiplier 0.9 gives the stated value.
A1
d=10 %.
A1

Final answer

10% per year.

Question 6

[4 marks]
Mina makes a deposit of 3650 US dollars earns a nominal annual interest rate of 4.8%, compounded monthly for two years. There are no later transactions. A nominal annual rate is divided by the number of compounding periods per year.
(a) State the monthly percentage rate and number of periods.
[2]
(b) Find the final balance to two decimal places.
[2]
Show solutionHide solution
(a) 4.8/12=0.4 % per month.
A1
N=24 months.
A1
(b) 3650(1+0.048/12)^{24}.
M1
4017.00 dollars.
A1

Final answer

(a) 0.4%, 24 periods. (b) USD 4017.00.

Question 7

[4 marks]
Jules’s ceramic studio finances a kiln with a 16 800-US-dollar loan. The nominal annual rate is 6.6%, compounded monthly. Equal repayments occur at each month end for 4 years and clear the loan. Use full precision internally.
Find the regular repayment, to two decimal places. State the TVM settings and cash-flow signs used.
[4]
Show solutionHide solution
N=48 payment periods.
A1
I\%=6.6, P/Y=C/Y=12; END.
A1
PV=16\,800, FV=0, solve PMT<0.
M1
Repayment 399.19 dollars.
A1

Final answer

USD 399.19; N=48, I\%=6.6, PV=16\,800, FV=0, PMT<0, P/Y=C/Y=12, END.

Question 8

[4 marks]
Lina’s fund has 22 400 US dollars invested at 0.4% monthly. It pays 800 US dollars immediately after each month-end interest credit while the fund can afford a full withdrawal.
Find the number of full withdrawals and the balance immediately after the last one. Justify with the next month's available balance. Give money to two decimal places.
[4]
Show solutionHide solution
Generate balances by B_{k+1}=1.004B_k-800.
M1
29 full withdrawals.
A1
After 29 withdrawals: 602.08 dollars.
A1
Next month available 604.48 dollars, below 800.
R1

Final answer

29 full withdrawals; USD 602.08 left. Next month USD 604.48 <800.

Question 9

[2 marks]
Sam’s model for weekly donations, in euros, is a positive geometric series that has first term 7.2 and infinite sum 48.
Find its common ratio and third term.
[2]
Show solutionHide solution
1-r=7.2/48=0.15, so r=0.85.
M1
u_3=7.2(0.85)^2=5.202.
A1

Final answer

r=0.85; u_3=5.202.

Question 10

[2 marks]
Tara’s archive compresses an image in successive passes. The first pass saves 6.8 MB. Each following pass saves 0.45 times as much space as the previous pass.
Find the maximum modelled saving. Can the saving reach 13 MB?
[2]
Show solutionHide solution
S_\infty=6.8/0.55=12.3636\ldots MB.
M1
It cannot reach 13 MB, since every finite sum is below 12.3636\ldots.
R1

Final answer

12.4 MB; the limit is below 13 MB.

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 11

[10 marks]
Mariam is following a training plan for a charity cycle ride. The distance she rides each week increases by the same number of kilometres from week to week. Over the first 10 weeks she rides a total of 195 km. The distance in week 10 is 3 km more than five times the distance in week 1.
(a) Find the distance she rides in week 1.
[3]
(b) Find the weekly increase in distance.
[2]
(c) Use your GDC to find the least number of weeks for which her total distance reaches 1000 km.
[3]
(d) Find the distance she would ride in the final week of part (c), and comment on whether the model is realistic over that many weeks.
[2]
Show solutionHide solution
(a) S_{10} = 5(u_1 + u_{10}) = 195 \Rightarrow u_1 + u_{10} = 39
[1]
u_{10} = 5u_1 + 3
[1]
6u_1 + 3 = 39 \Rightarrow u_1 = \mathbf{6} km.
[1]
(b) u_{10} = 33, so 6 + 9d = 33
[1]
d = \mathbf{3} km per week.
[1]
(c) S_n = \dfrac{n}{2}\bigl(12 + 3(n-1)\bigr) \ge 1000; table of S_n.
[1]
S_{24} = 972 and S_{25} = 1050.
[1]
So \mathbf{25} weeks.
[1]
(d) u_{25} = 6 + 24(3) = 78 km.
[1]
Riding 78 km in a single week of a plan that began at 6 km is not realistic for a beginner in a charity ride; the model would not continue to be valid.
[1]

Final answer

(a) 6 km. (b) 3 km per week. (c) 25 weeks. (d) 78 km; constant increases may exceed her available training time.

Question 12

[6 marks]
Nora and Dev’s two start-ups both launch an app in the same month. Company X has 2000 users in month 1 and gains 450 new users each month. Company Y has 1500 users in month 1 and its user count grows by 14% each month.
(a) Write down an expression for the number of users of each company in month n.
[2]
(b) Use your GDC to find the first month in which Company Y has more users than Company X.
[3]
(c) Company X's founder argues that her company is the stronger one because it is ahead for the first 13 months. Comment on this argument.
[1]
Show solutionHide solution
(a) X is arithmetic: x_n = 2000 + 450(n-1).
[1]
Y is geometric: y_n = 1500(1.14)^{n-1}.
[1]
(b) Table of x_n and y_n (or graph intersection).
[1]
Month 13: y_{13} = 7227 < x_{13} = 7400.
[1]
Month 14: y_{14} = 8239 > x_{14} = 7850. So month \mathbf{14}.
[1]
(c) A fixed amount added each month (linear growth) is eventually overtaken by a fixed percentage growth, whatever the start values; after month 14 the gap widens every month in Y's favour.
[1]

Final answer

(a) x_n=2000+450(n-1); y_n=1500(1.14)^{n-1}. (b) Month 14. (c) Later growth can reverse the early lead.

Question 13

[7 marks]
Leena models a solar panel. The total lifetime electricity output of a solar panel is modelled by an infinite geometric sequence of yearly outputs. The model predicts a total lifetime output of S_\infty = 50 MWh and a second-year output of 12 MWh.
(a) Show that the common ratio r satisfies 25r^2 - 25r + 6 = 0.
[2]
(b) Solve the equation in part (a) and hence find both possible values of the first-year output.
[2]
(c) The manufacturer states that the first-year output is more than 25 MWh. Find the third-year output.
[3]
Show solutionHide solution
(a) S_\infty = \dfrac{u_1}{1 - r} = 50 \Rightarrow u_1 = 50(1 - r).
[1]
u_2 = u_1 r = 50r(1 - r) = 12
[1]
50r - 50r^2 = 12 \Rightarrow 25r^2 - 25r + 6 = 0
AG
(b) (5r - 2)(5r - 3) = 0, so r = 0.4 or r = 0.6; both satisfy |r| < 1.
[2]
(c) r = 0.4: u_1 = 50(0.6) = 30; r = 0.6: u_1 = 50(0.4) = 20. The first-year output is above 25, so r = 0.4.
[2]
u_3 = 12 \times 0.4 = \mathbf{4.8} MWh.
[1]

Final answer

(a) u_1=50(1-r), u_1r=12. (b) r=0.4 or 0.6; u_1=30 or 20 MWh respectively. (c) 4.8 MWh.

Question 14

[8 marks]
Ivo observes a garden snail that crawls 90 cm in the first hour of its route along a straight path. In each later hour it crawls 85% of the distance it crawled in the previous hour.
(a) Show that the snail will never get further than 6 metres from its starting point, however long it crawls.
[2]
(b) Use your GDC to find the first hour by the end of which the snail has crawled more than 5 metres.
[3]
(c) Find the least number of hours after which the snail is within 1 cm of the point it is approaching.
[3]
Show solutionHide solution
(a) Distances form a geometric sequence, u_1 = 90, r = 0.85, |r| < 1.
[1]
S_\infty = \dfrac{90}{1 - 0.85} = 600 cm = 6 m, and every partial sum is below the limit.
[1]
AG
(b) S_n = \dfrac{90(1 - 0.85^n)}{0.15} > 500; table.
[1]
S_{11} = 499.6\ldots < 500 and S_{12} = 514.7\ldots > 500.
[1]
So hour \mathbf{12}.
[1]
(c) S_\infty - S_n = \dfrac{90 \times 0.85^n}{0.15} = 600(0.85)^n < 1
[1]
600(0.85)^{39} = 1.06\ldots > 1 and 600(0.85)^{40} = 0.90\ldots < 1.
[1]
So \mathbf{40} hours.
[1]

Final answer

(a) Limit 600 cm =6 m; every finite sum is below it. (b) Hour 12. (c) 40 hours.

Question 15

[7 marks]
Paulo borrows NZD 26 500 to buy a caravan. The nominal annual interest rate is 5.7%, compounded monthly, and the loan is repaid in equal monthly instalments over 6 years.
(a) Find Paulo's monthly instalment.
[2]
(b) Find the amount Paulo still owes after he has made 30 payments.
[3]
(c) Find the total interest that Paulo has paid during the first 30 payments.
[2]
Show solutionHide solution
(a) N=72, I\%=5.7, PV=26\,500, FV=0, P/Y=C/Y=12, END. The stored payment is -435.4385\ldots, so Paulo pays NZD 435.44 a month (M1 A1). (b) Keep the stored payment and original principal, set N=30, and solve for the balance (M1 A1). The balance is NZD 16 544.17 (A1). (c) Principal repaid is 26\,500-16\,544.1696\ldots NZD. Interest is 30(435.4385\ldots)-(26\,500-16\,544.1696\ldots)=3107.33 NZD (M1 A1).

Final answer

(a) NZD 435.44. (b) NZD 16 544.17. (c) NZD 3107.33.

Question 16

[7 marks]
Isabel already has USD 5000 in an account that pays 4.5% per year compounded monthly. She decides to deposit a further USD 600 at the end of every month to save for a car costing USD 50 000.
(a) Use your GDC to find how many monthly deposits are needed for the account to reach USD 50 000. Write down the values you enter.
[3]
(b) Hence find the least number of whole months, and express this in years and months.
[2]
(c) Find the balance immediately after the 64th deposit, and the balance immediately after the 65th deposit, each to the nearest dollar.
[2]
Show solutionHide solution
(a) Enter: N=?, I\%=4.5, PV=-5000, PMT=-600, FV=50\,000, P/Y=12, C/Y=12
[2]
N = 64.43\ldots
[1]
(b) Deposits are made at month ends, so 65 deposits are needed (N=64.4 means that the target is first reached at the 65th).
[1]
65 months = 5 years 5 months.
[1]
(c) After 64: N = 64: FV = \$49\,662.
[1]
After 65: FV = \$50\,448.
[1]

Final answer

(a) N=64.43\ldots. (b) 65 months; 5 years 5 months. (c) USD 49 662; USD 50 448.

Question 17

[7 marks]
Hugo manages a trust fund of £500 000 pays £3200 to a charity at the end of every month. The fund earns a nominal annual interest rate of 5.1%, compounded monthly.
(a) Find the amount in the fund after 10 years.
[3]
(b) Find the total interest that the fund earns during the 10 years.
[2]
(c) Find the interest the fund earns in the first month. Hence state, with a reason, whether the amount in the fund is rising or falling.
[2]
Show solutionHide solution
(a) The fund is paid in (negative PV) and the charity receives the payments (positive PMT):
N = 120,\ I\% = 5.1,\ PV = -500\,000,\ PMT = +3200,\ FV = ?,\ P/Y = C/Y = 12
FV = 332\,175.4252.
[2]
The fund holds \textbf{\text{£}332\,175.43}.
[1]
(b) Paid out = 120 \times 3200 = 384\,000.00, and the fund has fallen by 500\,000 - 332\,175.43 = 167\,824.57.
[1]
Interest = 384\,000.00 - 167\,824.57 = \textbf{\text{£}216\,175.43}.
[1]
(c) First month's interest = 0.00425 \times 500\,000 = \textbf{\text{£}2125.00}.
[1]
This is less than the £3200 withdrawn, so the fund is falling.
[1]

Final answer

(a) GBP 332 175.43. (b) GBP 216 175.43. (c) GBP 2125.00 <3200, so it falls.

Question 18

[10 marks]
Carla borrows USD 15 000 at 7.2% per year compounded monthly. She repays USD 320 at the end of every month until the loan is cleared; the last payment may be smaller than USD 320.
(a) Use your GDC to find how many months it takes to repay the loan. Write down the values you enter.
[3]
(b) Find the amount still owed after 55 payments and hence the final (56th) payment, to the nearest cent.
[4]
(c) Find the total amount Carla pays and the interest she pays.
[3]
Show solutionHide solution
(a) Enter: N=?, I\%=7.2, PV=15\,000, PMT=-320, FV=0, P/Y=12, C/Y=12
[2]
N = 55.21\ldots, so 56 payments are needed (the 56th is smaller).
[1]
(b) After 55 payments: N=55, I\%=7.2, PV=15\,000, PMT=-320, FV=?, P/Y=12, C/Y=12 gives FV = -65.44, so the balance is \$65.44.
[2]
One more month of interest: 65.44 \times 1.006 = \$65.83 is the final payment.
[2]
(c) Total = 55 \times 320 + 65.83 = \$17\,665.83.
[2]
Interest = 17\,665.83 - 15\,000 = \$2\,665.83.
[1]

Final answer

(a) 56 months. (b) USD 65.44; USD 65.83. (c) USD 17 665.83 total; USD 2665.83 interest.

Question 19

[12 marks]
Amir saves for retirement. At the end of each year for 18 years he pays USD 2400 into a fund earning 4.6% per year, compounded annually. He then stops paying in and, starting one month later, withdraws a fixed amount USD W at the end of each month for 20 years. During the withdrawal period the fund earns 3.0% per year compounded monthly, and the fund is exactly empty after the last withdrawal.
(a) Find the value of the fund at the end of the 18 years, to the nearest dollar.
[2]
(b) Find the value of W, to the nearest cent.
[3]
(c) Find the total amount Amir pays in and the total amount he withdraws, and find the difference between them.
[3]
(d) Amir would like to withdraw at least USD 1200 per month. Use your GDC to find the least yearly payment he should make during the 18 years, to the nearest dollar, if the other conditions are unchanged.
[4]
Show solutionHide solution
(a) Enter: N=18, I\%=4.6, PV=0, PMT=-2400, FV=?, P/Y=1, C/Y=1
[1]
FV = \$65\,052 (65 052.03).
[1]
(b) Withdrawal stage: N=240, I\%=3, PV=-65\,052.03, PMT=?, FV=0, P/Y=12, C/Y=12 (keep the unrounded fund value)
[2]
PMT = 360.78, so W = \$360.78.
[1]
(c) Pays in: 18 \times 2400 = \$43\,200.
[1]
Withdraws: 240 \times 360.777\ldots = \$86\,586.
[1]
Difference \approx \$43\,386 is the interest earned over the 38 years.
[1]
(d) The fund needed: N=240, I\%=3, PV=?, PMT=1200, FV=0, P/Y=12, C/Y=12; PV = -216\,373.10, so the fund must be worth at least \$216 373 at retirement.
[2]
Savings stage: N=18, I\%=4.6, PV=0, PMT=?, FV=216\,373.1, P/Y=1, C/Y=1 gives PMT = -7\,982.77, so the least whole-dollar payment is \mathbf{\$7983} per year.
[2]

Final answer

(a) USD 65 052. (b) USD 360.78. (c) USD 43 200 in, USD 86 586 out; USD 43 386 difference. (d) USD 7983 per year.

Question 20

[9 marks]
Nadia’s rubber ball is dropped onto a concrete floor from a height of 3.2 m. Each time it bounces, it rises to 70% of the height from which it fell. (a) Find the greatest height the ball reaches after its 4th bounce.
[2]
(b) Find how far the ball travels up and down, in total, between being dropped and hitting the floor for the 6th time.
[3]
(c) Find the number of the first bounce after which the ball rises less than 10 cm.
[2]
(d) Nadia says the ball will travel less than 20 m in total, however long it keeps bouncing. Determine whether she is right.
[2]
Show solutionHide solution
(a) 3.2(0.7)^4=0.76832 m, hence 0.768 m (M1 A1). (b) First drop once and five rebounds twice: 3.2+2[2.24(1-0.7^5)/0.3]=15.623488 m, hence 15.6 m (M1 M1 A1). (c) 3.2(0.7)^k<0.1; heights at bounces 9 and 10 are 0.129 m and 0.0904 m, so bounce 10 (M1 A1). (d) Since 0<0.7<1, the limiting distance is 3.2+2(2.24)/0.3=18.1333\ldots m, less than 20 m (M1). Every finite path is below this limit, so Nadia is right within the model (R1).

Final answer

(a) 0.768 m. (b) 15.6 m. (c) Bounce 10. (d) Yes: limiting distance 18.1 m <20 m.

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 21

[12 marks]
Zara is buying a car for USD 32 000 and compares two finance offers. Both are repaid by equal monthly instalments, with interest compounded monthly.
Offer A: a 20% deposit, and the rest repaid over 4 years at 4.8% per year.
Offer B: a USD 2000 deposit, and the rest repaid over 5 years at 3.9% per year.
(a) Find the amount borrowed and the monthly instalment under Offer A, to the nearest cent.
[3]
(b) Find the amount borrowed and the monthly instalment under Offer B, to the nearest cent.
[2]
(c) Find the total cost of the car (deposit plus all instalments) under each offer, and state which is cheaper and by how much.
[4]
(d) Zara can afford at most USD 560 per month. Which offer(s) can she choose? Comment on whether the cheaper offer is the better one for her.
[3]
Show solutionHide solution
(a) Deposit is 0.20(32\,000)=6400 USD; principal is USD 25 600 (A1). With N=48, I\%=4.8, FV=0, P/Y=C/Y=12, END, the monthly payment is USD 587.23 (M1 A1). (b) Principal is 32\,000-2000=30\,000 USD; with N=60 and I\%=3.9 the monthly payment is USD 551.14 (M1 A1). (c) Retain both full-precision payments. A costs 6400+48(587.233422\ldots)=34\,587.20 USD (M1 A1). B costs 2000+60(551.142868\ldots)=35\,068.57 USD (A1). A costs USD 481.37 less (A1). (d) A needs more than USD 560 monthly and B needs less (A1 A1). Only B fits Zara’s monthly budget, even though A is cheaper overall (R1).

Final answer

(a) USD 25 600; USD 587.23. (b) USD 30 000; USD 551.14. (c) USD 34 587.20; USD 35 068.57; A is cheaper by USD 481.37. (d) Only B fits the USD 560 budget.

Question 22

[11 marks]
Ellis’s small bakery takes out a business loan of USD 50 000 at a nominal annual interest rate of 7.8%, compounded monthly, repaid in equal monthly instalments over 6 years.
(a) Find the monthly repayment.
[3]
(b) Find the total interest paid over the 6 years.
[2]
(c) Find the outstanding balance immediately after the payment at the end of year 3.
[3]
(d) From the start of the loan the owner instead pays USD 100 more than the required repayment each month. Find how many months it now takes to repay the loan.
[3]
Show solutionHide solution
(a) N=72, I\%=7.8, PV=50\,000, FV=0, P/Y=C/Y=12, END gives a stored monthly payment of 871.7873\ldots USD, hence USD 871.79 (M1 A1 A1). (b) With the stored payment, total interest is 72(871.7873\ldots)-50\,000=12\,768.69 USD (M1 A1). If all regular payments are rounded to USD 871.79, the product gives USD 12 768.88; this uses a different convention. (c) Set N=36 and retain the stored payment; solve for the balance (M1 A1), USD 27 902.39 (A1). (d) With a payment 100 USD larger, solve for the term, giving 62.8\ldots months (M1 A1). It takes 63 months, including a smaller final payment (A1).

Final answer

(a) USD 871.79. (b) USD 12 768.69 with the stored payment. (c) USD 27 902.39. (d) 63 months.

Question 23

[11 marks]
The Fernandez family takes out a mortgage of USD 320 000 at a nominal annual interest rate of 3.9%, compounded monthly, repaid in equal monthly instalments over 30 years.
(a) Find the monthly repayment.
[3]
After exactly 10 years the family refinances: the outstanding balance is transferred to a new loan at a nominal annual rate of 3.2%, compounded monthly, to be repaid over the next 20 years.
(b) Find the outstanding balance at the moment they refinance.
[3]
(c) Find the new monthly repayment under the refinanced loan.
[2]
(d) By comparing the total still to be paid under each option, find how much the family saves over the remaining 20 years by refinancing.
[3]
Show solutionHide solution
(a) N=360, I\%=3.9, PV=320\,000, FV=0, P/Y=C/Y=12, END gives USD 1509.34 per month (M1 A1 A1). (b) Retain the payment and use N=120 with the original principal (M1 A1). The outstanding balance is USD 251 253.20 (A1). (c) This unrounded balance becomes the principal of a 240-month loan at I\%=3.2, giving USD 1418.73 monthly (M1 A1). (d) Compare 240 old payments with 240 new ones (M1). Using both full-precision payments gives USD 21 745.16 saved (A1 A1). Multiplying the displayed cent-rounded payments instead gives USD 21 746.40.

Final answer

(a) USD 1509.34. (b) USD 251 253.20. (c) USD 1418.73. (d) USD 21 745.16 using full precision.

Question 24

[10 marks]
A simplified model describes the medicine given to patient Alex. Immediately after each dose, the amount of drug in the bloodstream (in mg) follows a repeatable pattern. A dose of 20 mg is given every 6 hours. During each 6-hour interval, the body eliminates 30% of the drug present, so that just before the next dose, 70% of the previous post-dose amount remains.
Let A_n be the amount of drug in the bloodstream immediately after the nth dose.
(a) Show that A_1 = 20 and find A_2 and A_3.
[3]
(b) Explain why A_{n+1} = 0.7A_n + 20.
[2]
(c) Using a GDC (or otherwise), find A_6, the amount immediately after the 6th dose.
[2]
(d) The long-run amount just after a dose settles towards a limit L satisfying L = 0.7L + 20. Find L.
[3]
Show solutionHide solution
(a) The first dose gives A_1=20 mg; A_2=0.7(20)+20=34 mg; A_3=0.7(34)+20=43.8 mg (A1 A1 A1). (b) Multiply by 0.7 for the amount remaining (M1), then add the new dose of 20 mg (R1). (c) Iteration gives A_4=50.66, A_5=55.462, A_6=58.8234 mg, hence 58.8 mg to 3 significant figures (M1 A1). (d) At a fixed point L=0.7L+20 (M1), so 0.3L=20 (M1), giving L=66.7 mg to 3 significant figures (A1). This simplified model assumes a fixed eliminated fraction.

Final answer

(a) 20, 34, 43.8 mg. (b) A_{n+1}=0.7A_n+20. (c) 58.8 mg. (d) 66.7 mg.

Question 25

[9 marks]
Sione budgets NZD 9000 at today's prices for a trip in five years. Prices rise by 3.1% annually. His account pays 4.5% nominal interest per year, compounded monthly.
(a) Find the trip's expected cost in five years.
[2]
(b) Find the single deposit needed today to meet that cost.
[2]
(c) Instead Sione deposits NZD 1000 today and equal amounts at each month end. Find the monthly amount needed.
[3]
(d) Find the interest earned in the plan in part (c).
[2]
Show solutionHide solution
(a) 9000(1.031)^5=10\,484.21 NZD (M1 A1). (b) N=60, I\%=4.5, PMT=0, FV=10\,484.21\ldots, P/Y=C/Y=12, END gives a present investment of NZD 8375.34 (M1 A1). (c) Set PV=-1000, retaining the unrounded target. The monthly deposit is NZD 137.50 (M1 A1 A1). (d) 10\,484.21\ldots-1000-60(137.4986\ldots)=1234.30 NZD interest (M1 A1). The model assumes unchanged rates and no account fees.

Final answer

(a) NZD 10 484.21. (b) NZD 8375.34. (c) NZD 137.50. (d) NZD 1234.30.

Question 26

[18 marks]
Ravi manages a solar farm that produced 2600 MWh of electricity in its first year of operation. The output is modelled to fall by 0.9% each year. (a) Find the output in year 12.
[2]
(b) Find the total output over the first 25 years.
[2]
(c) Show that, according to the model, the total output over the farm's whole life can never exceed 290 000 MWh.
[2]
The electricity is sold for EUR 92 per MWh. The farm was built with a loan of EUR 2 400 000 at a nominal annual interest rate of 4.8%, compounded monthly, repaid by equal monthly instalments over 15 years. (d) Find the monthly instalment.
[3]
(e) Find the first year in which the income from selling the farm's electricity is less than the total of the instalments paid in that year.
[3]
Immediately after the instalment at the end of year 8, the owners repay the rest of the loan in a single payment. (f) Find this payment.
[3]
(g) Find the amount the owners save by settling the loan early.
[2]
(h) Suggest one reason why the model for the output may not be suitable for the whole life of the farm.
[1]
Show solutionHide solution
(a) 2600(0.991)^{11}=2350 MWh to 3 significant figures (M1 A1). (b) 2600(1-0.991^{25})/0.009=58\,400 MWh to 3 significant figures (M1 A1). (c) |r|=0.991<1 gives S_\infty=2600/0.009=288\,888.888\ldots MWh (M1). Every finite total is below this limit, which is less than 290 000 MWh (R1). (d) N=180, I\%=4.8, PV=2\,400\,000, FV=0, P/Y=C/Y=12, END gives EUR 18 729.95 monthly (M1 A1 A1). (e) Income in year n is 239\,200(0.991)^{n-1} EUR; annual payments total EUR 224 759.36 using the stored payment (M1). In years 7 and 8 income is EUR 226 570.36 and EUR 224 531.23 (A1), so year 8 is the first shortfall (A1). (f) Set N=96 with the original principal and stored payment; the balance is EUR 1 334 030.06 (M1 A1 A1). (g) 84(18\,729.946\ldots)-1\,334\,030.06\ldots=239\,285.44 EUR saved (M1 A1). (h) Panels can fail or be replaced, so a constant 0.9% annual fall is unlikely throughout the farm’s life (R1).

Final answer

(a) 2350 MWh. (b) 58 400 MWh. (c) Limit 288\,888.888\ldots<290\,000 MWh. (d) EUR 18 729.95. (e) Year 8. (f) EUR 1 334 030.06. (g) EUR 239 285.44. (h) Panels may fail or be replaced.

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 A loan is a recurrence

G1 Worked

[7 marks]
Nico borrows EUR 12000 at 0.5\% per month and repays EUR 250 at each month end.
(a) Find his balance after 12 payments using the recurrence formula.
[3]
(b) Find the number of payments needed to clear the loan.
[2]
(c) Find the interest-only monthly payment.
[2]
Show solutionHide solution
(a) i=0.005, P=250, so P/i=50000 (M1). With B_0=12000, B_k=50000-38000(1.005)^k (A1), giving EUR 9656.24 after 12 payments (A1). (b) Solve 1.005^k=50000/38000 to get k=55.02\ldots (M1); therefore 55 full payments and a smaller 56th are needed (A1). (c) The boundary is iB_0=60 EUR (M1). At that payment only interest is covered; a payment greater than EUR 60 clears the loan (R1).

Final answer

EUR 9656.24; 56 payments; EUR 60 boundary.

G2 Paying for a stream of prizes

G2 Investigation

[21 marks]
Elena organises a town music prize of EUR 2000, paid at the end of every year. The fund earns 4\% per year, compounded annually. Assume the rate stays fixed and there are no fees.
(a) Show that the amount needed today for the prize at the end of year k is 2000(1.04)^{-k}. Hence show that the first n prizes cost 50000(1-1.04^{-n}) today.
[4]
(b) Find the cost for 20 prizes and check it with the finance solver. Find the cost of prizes for ever. Then find the least n for which the finite fund costs at least 90\% of the perpetual fund.
[7]
(c) The first prize remains EUR 2000 but each later prize is 2\% larger. Show that the required fund for ever is EUR 100000. Explain what changes if the prize grows by 4\% or more each year.
[5]
(d) Instead, the town invests EUR 50000 and pays a fixed EUR 2500 each year. If B_k is the balance after k prizes, show that B_k=62500-12500(1.04)^k.
[3]
(e) Find the number of full EUR 2500 prizes this fund can pay, with a neighbour check.
[2]
Show solutionHide solution
(a) Discounting EUR 2000 for k years gives 2000(1.04)^{-k} (A1). The present values have first term 2000/1.04 and ratio 1/1.04 (M1). Their finite sum is (2000/1.04)(1-1.04^{-n})/(1-1/1.04) (M1), which simplifies to 50000(1-1.04^{-n}) (A1). (b) At n=20 this is EUR 27180.65; N=20, I\%=4, PMT=2000, FV=0, P/Y=C/Y=1, END gives PV=-27180.65\ldots (M1 A1). Since 1/1.04<1, the perpetual fund is EUR 50000 (M1 A1). For 90 percent, solve 1.04^{-n}\le0.1 (M1). The continuous boundary is 58.71\ldots years; the 58-year fraction is below 90 percent and the 59-year fraction above it (A1), so 59 years are needed (A1). (c) The discounted first term is 2000/1.04 and the ratio is 1.02/1.04 (M1). The sum is (2000/1.04)/(1-1.02/1.04)=100000 EUR (M1 A1). With growth at least 4 percent, the discounted ratio is at least one (M1), so no finite initial fund suffices under the model (R1). (d) The recurrence is B_{k+1}=1.04B_k-2500 with B_0=50000 (M1). Subtract the fixed point 62500 (M1). Then B_k-62500=(50000-62500)1.04^k, as required (A1). (e) Solve B_k=0, giving k=41.04\ldots (M1). After 41 prizes the balance is EUR 86.73, while the algebraic balance after 42 prizes is negative. Therefore the fund pays 41 full prizes (A1).

Final answer

EUR 27180.65; EUR 50000; 59 years; EUR 100000; 41 prizes.

From the book

Photon Guidebook: Mathematics Applications and Interpretation HL for the IB Diploma

Chapter 2 teaches sequences, series and financial mathematics from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book

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