Free worked solutions · 2026 Edition

IB Maths AI SL 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 SL 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]
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]
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, height 61.2 cm

Question 3

[2 marks]
A display contains 18 strips. The number of tiles on each strip is arithmetic, from 23 on the first to 74 on the last.
Find the total number of tiles.
[2]
Show solutionHide solution
S_{18}=18(23+74)/2
M1
S_{18}=873 tiles
A1

Final answer

873 tiles

Question 4

[4 marks]
A geometric sequence 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 5

[3 marks]
A coating 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 6

[3 marks]
A print press falls in value from 15 000 dollars to 10 935 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 7

[4 marks]
A deposit of 3650 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% per month, 24 periods
  • (b) 4017.00 dollars

Question 8

[4 marks]
A ceramic studio finances a kiln with a 16 800-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

399.19 dollars

Question 9

[4 marks]
A chess club saves for travel. Starting with zero, it deposits 135 dollars at each month end for 30 periods into an account paying a nominal annual rate of 3.6%, compounded monthly. Value the fund immediately after the final deposit.
Find the fund value and the interest earned, to two decimal places.
[4]
Show solutionHide solution
N=30, I\%=3.6, PV=0, PMT=-135, P/Y=C/Y=12; END, solve FV>0.
M1
Fund 4231.21 dollars.
A1
Interest =FV-30(135).
M1
181.21 dollars.
A1

Final answer

fund 4231.21 dollars; interest 181.21 dollars

Question 10

[4 marks]
A fund has 22 400 dollars invested at 0.4% monthly. It pays 800 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; 602.08 dollars left (the next month's 604.48 dollars is below 800)

Tutorial

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

Question 11

[6 marks]
The depth of water in a reservoir is measured at the start of each week, and it falls by the same number of centimetres every week. At the start of week 3 the depth is 412 cm and at the start of week 9 the depth is 340 cm.
(a) Find the weekly fall in depth.
[2]
(b) Find the depth at the start of week 1, and write down a formula for the depth u_n at the start of week n.
[2]
(c) Use your GDC to find the first week in which the depth is below 250 cm.
[2]
Show solutionHide solution
(a) Six weeks separate u_3 and u_9: 6d = 340 - 412 = -72.
M1[1]
d = -12, so the depth falls by \mathbf{12} cm each week.
A1[1]
(b) u_1 = 412 + 2(12) = \mathbf{436} cm.
A1[1]
u_n = 436 - 12(n-1) = 448 - 12n.
A1[1]
(c) Solve 448 - 12n < 250 (table or solver): n > 16.5.
M1[1]
u_{16} = 256 and u_{17} = 244, so the first week is week \mathbf{17}.
A1[1]

Final answer

  • (a) 12 cm
  • (b) 436 cm, u_n=448-12n
  • (c) week 17

Question 12

[5 marks]
A paved courtyard is built around a square fountain. The first ring of paving stones around the fountain contains 20 stones, and ring k contains 8k + 12 stones, for k = 1, 2, 3, \ldots
(a) Show that the numbers of stones in the rings form an arithmetic sequence and state the common difference.
[2]
(b) The courtyard has 15 rings. Find the total number of stones used, and write this total using sigma notation.
[3]
Show solutionHide solution
(a) u_{k+1} - u_k = \bigl(8(k+1) + 12\bigr) - (8k + 12) = 8, which is constant, so the sequence is arithmetic.
R1[1]
The common difference is d = 8.
A1[1]
(b) The total is \displaystyle\sum_{k=1}^{15}(8k + 12).
A1[1]
u_{15} = 132, so S_{15} = \dfrac{15}{2}(20 + 132)
M1[1]
= \mathbf{1140} stones (the GDC sum template gives the same value).
A1[1]

Final answer

  • (a) d=8
  • (b) \sum_{k=1}^{15}(8k+12)=1140 stones

Question 13

[6 marks]
A rainwater tank holds 8000 litres at the start of day 1. A slow leak means that the volume in the tank falls by 5% each day (no rain falls and no water is used).
(a) Find the volume in the tank at the start of day 8, to 3 significant figures.
[2]
(b) Find the first day on which the volume at the start of the day is below 5000 litres.
[2]
(c) Find the first day on which the volume at the start of the day is less than half of the original 8000 litres.
[2]
Show solutionHide solution
(a) Each day keeps 95%, so r = 0.95 and u_8 = 8000(0.95)^7
M1[1]
= 5586.7\ldots \approx \mathbf{5590} litres.
A1[1]
(b) Table of 8000(0.95)^{n-1}: u_{10} = 5042 > 5000 and u_{11} = 4790 < 5000.
M1[1]
The first day is day \mathbf{11}.
A1[1]
(c) Solve 8000(0.95)^{n-1} < 4000 (table or graph): u_{14} = 4107 > 4000 and u_{15} = 3901 < 4000.
M1[1]
The first day is day \mathbf{15}.
A1[1]

Final answer

  • (a) 5590 litres
  • (b) day 11
  • (c) day 15

Question 14

[8 marks]
Isla is offered a 20-day job and can choose how she is paid.
Option 1: $60 on day 1, and each day $45 more than the day before.
Option 2: $5 on day 1, and each day 40% more than the day before.
(a) Find the total Isla is paid over the 20 days under Option 1.
[2]
(b) Find the total Isla is paid over the 20 days under Option 2, to the nearest dollar.
[3]
(c) Find the first day on which Isla's total pay under Option 2 is greater than her total pay under Option 1. Show the totals for the day before.
[3]
Show solutionHide solution
(a) Option 1 is arithmetic (u_1 = 60, d = 45): S_{20} = \dfrac{20}{2}\bigl(2(60) + 19(45)\bigr)
M1[1]
= \mathbf{\$9750}.
A1[1]
(b) Option 2 is geometric (u_1 = 5, r = 1.4): S_{20} = \dfrac{5(1.4^{20} - 1)}{1.4 - 1} M1A1
[2]
= 10\,446.03\ldots \approx \mathbf{\$10\,446}.
A1[1]
(c) Table of both totals on the GDC:
M1[1]
day 19: Option 1 = \$8835, Option 2 = \$7458 (Option 1 still ahead).
A1[1]
day 20: Option 1 = \$9750, Option 2 = \$10\,446. So the first day is day \mathbf{20}.
A1[1]

Final answer

  • (a) USD 9750
  • (b) USD 10 446
  • (c) day 20 (day 19: 8835 against 7458)

Question 15

[7 marks]
Yusuf borrows $25 000 at a nominal annual interest rate of 9%, compounded monthly, and repays it with equal monthly instalments over 5 years.
(a) Find Yusuf's monthly instalment.
[2]
(b) Find the amount still owed after 30 instalments, and state whether this is more or less than half of the loan.
[2]
(c) Find the smallest number of instalments after which the amount owed is less than half of the loan. Show how you obtained your answer.
[3]
Show solutionHide solution
(a)
N = 60, I\% = 9, PV = +25\,000, PMT = ?, FV = 0, P/Y = C/Y = 12
Solving for PMT gives −518.9589, so the instalment is \textbf{\$518.96}. M1A1
[2]
(b) 30 instalments remain: solving for PV with N = 30, PMT = -518.9589, FV = 0 gives 13 895.1657.
M1[1]
The balance is \textbf{\$13\,895.17}, which is more than half of the loan ($12 500).
A1[1]
(c) Find the number of instalments N that a balance of $12 500 can still support: PV = +12\,500, PMT = -518.9589, FV = 0; solving for N gives 26.6654.
M1[1]
So the balance is below $12 500 once the number of instalments remaining is less than 26.67, i.e. after k > 60 - 26.67 = 33.33 instalments.
M1[1]
The smallest number is \textbf{34} (balances: $12 641.58 after 33, $12 217.43 after 34).
A1[1]

Final answer

  • (a) USD 518.96
  • (b) USD 13 895.17, more than half
  • (c) 34 instalments

Question 16

[7 marks]
Gabriel wants to have $18 000 for a wedding in 4 years' time. He will pay equal amounts into an account at the end of every month. The account pays a nominal annual interest rate of 2.4%, compounded monthly, and starts with a balance of zero.
(a) Find the monthly payment Gabriel needs to make.
[3]
(b) Find the total amount Gabriel pays in, and the interest earned.
[2]
Gabriel decides he can only afford $340 per month.
(c) Find the number of whole months after which his account first holds at least $18 000.
[2]
Show solutionHide solution
(a) Solver entries: N=48, I\%=2.4, PV=0, PMT=?, FV=18\,000, P/Y=C/Y=12. M1A1
[2]
PMT = -357.66, so Gabriel pays \mathbf{\$357.66} per month (the negative sign shows money paid out).
A1[1]
(b) Total paid in = 48 \times 357.6625 = \mathbf{\$17\,167.80} (using the unrounded payment).
M1[1]
Interest = 18\,000 - 17\,167.80 = \mathbf{\$832.20}.
A1[1]
(c) Solver entries: N=?, I\%=2.4, PV=0, PMT=-340, FV=18\,000, P/Y=C/Y=12, giving N = 50.37.
M1[1]
Round up to a whole number of payments: after 50 months the balance is only $17 860.29, and after 51 months it is $18 236.01. The first whole number of months is \mathbf{51}.
A1[1]

Final answer

  • (a) USD 357.66
  • (b) USD 17 167.80 paid in, USD 832.20 interest
  • (c) 51 months

Question 17

[7 marks]
A university fund holds $165 000 in an account paying a nominal annual interest rate of 3.9%, compounded monthly. At the end of every month a scholarship of $1500 is paid out of the fund.
(a) Find the number of full scholarship payments of $1500 that the fund can make.
[3]
(b) After the last full payment, the fund holds a small balance. Find the balance, and the size of the final smaller payment that empties the fund one month later.
[2]
(c) The university would like the fund to last exactly 10 years. Find the monthly payment that does this.
[2]
Show solutionHide solution
(a) Solver entries: N=?, I\%=3.9, PV=-165\,000, PMT=1500, FV=0, P/Y=C/Y=12 (money paid into the fund is negative; money paid out is positive). M1A1
[2]
N = 136.34, so there are \mathbf{136} full payments (the 137th would not be possible in full).
A1[1]
(b) Entering N = 136 and solving for FV gives FV = 509.79: the balance is \mathbf{\$509.79}.
M1[1]
One month later this has grown by 0.325%, so the final payment is 509.79 \times 1.00325 = \mathbf{\$511.45}.
A1[1]
(c) Solver entries: N=120, I\%=3.9, PV=-165\,000, PMT=?, FV=0, P/Y=C/Y=12.
M1[1]
PMT = \mathbf{\$1662.71} per month.
A1[1]

Final answer

  • (a) 136 full payments
  • (b) USD 509.79; final payment USD 511.45
  • (c) USD 1662.71

Question 18

[7 marks]
A 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]
Show solutionHide solution
(a) 3.2(0.7)^4 M1 =0.76832 m, so 0.768 m A1 (b) The drop of 3.2 m counts once. The five rebounds before the 6th contact start at 3.2\times0.7=2.24 m with ratio 0.7, and each is travelled up and down: 3.2+2\cdot\frac{2.24(1-0.7^5)}{0.3} M1M1 =15.623\ldots, so 15.6 m A1 (c) 3.2(0.7)^k<0.1: k=9 gives 0.129 m and k=10 gives 0.0904 m M1 Bounce 10
A1

Final answer

  • (a) 0.768 m
  • (b) 15.6 m
  • (c) bounce 10

Question 19

[6 marks]
A drone inspects a straight pipeline that runs east to west through its base B. On each flight it flies from B to a turning point and straight back to B. The turning point of flight n is x_n metres east of B, where a negative value means west of B. The values x_1, x_2, x_3, \dots form a geometric sequence with x_1 = 50 and x_2 = -60.
(a) Write down the common ratio.
[1]
(b) Find the position of the turning point of flight 5.
[2]
(c) Find the total distance the drone flies in its first 6 flights.
[3]
Show solutionHide solution
(a) r=\frac{-60}{50}=-1.2 A1 (b) x_5=50(-1.2)^4 M1 =103.68, so 103.68 m east of B A1 (c) The distances out are |x_n|=50(1.2)^{n-1}, because distance ignores direction M1 Each flight is out and back: 2\times\frac{50(1.2^6-1)}{1.2-1} M1 =992.992, so 993 m
A1

Final answer

  • (a) −1.2
  • (b) 103.68 m east of B
  • (c) 993 m

Question 20

[6 marks]
A farmer buys a tractor for $62 000. After 4 years it is worth $41 500. Its value falls by the same percentage each year.
(a) Find the annual rate of depreciation, as a percentage to 3 significant figures.
[3]
(b) Find the value of the tractor after 10 years, to the nearest dollar.
[1]
(c) Find the number of whole years after which the tractor is worth less than 10% of its original price.
[2]
Show solutionHide solution
(a) 62\,000\,r^{4} = 41\,500, so r^4 = \dfrac{41\,500}{62\,000} = 0.66935\ldots
M1[1]
r = 0.90451\ldots (or solver: N=4, I\%=?, PV=-62\,000, PMT=0, FV=41\,500, P/Y=C/Y=1, giving I\% = -9.5488).
A1[1]
The annual rate of depreciation is \mathbf{9.55\%}.
A1[1]
(b) 62\,000\,(0.904512)^{10} \approx \mathbf{\$22\,727} (using the unrounded ratio).
A1[1]
(c) Solve 62\,000\,r^n < 6200 with the unrounded r (table or graph): n > 22.94.
M1[1]
n = 22 gives $6815.57 (still above $6200) and n = 23 gives $6164.76, so \mathbf{23} years.
A1[1]

Final answer

  • (a) 9.55%
  • (b) USD 22 727
  • (c) 23 years

Mastery

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

Question 21

[14 marks]
Elif is buying a car with a cash price of £32 000. The dealer offers two options, each with no deposit and equal monthly instalments over 5 years.
Option 1: finance from the dealer at a nominal annual interest rate of 2.9%, compounded monthly.
Option 2: a cash rebate of £2500 off the price, with the remaining price borrowed from a bank at a nominal annual interest rate of 6.4%, compounded monthly.
(a) Find the monthly instalment and the total repaid under Option 1.
[3]
(b) Find the monthly instalment and the total repaid under Option 2.
[3]
(c) State which option has the lower total repaid, and by how much.
[2]
(d) Find the size of the rebate, to the nearest pound, at which the two options would have the same total repaid.
[3]
(e) Elif can afford to pay at most £600 per month. Determine which option(s) she can afford.
[2]
(f) State one factor, other than the total repaid, that Elif should consider before choosing.
[1]
Show solutionHide solution
(a)
N = 60, I\% = 2.9, PV = +32\,000, PMT = ?, FV = 0, P/Y = C/Y = 12
Solving for PMT gives −573.5772, so the instalment is \textbf{\text{£}573.58}. M1A1
[2]
Total repaid = 60 \times 573.58 = \textbf{\text{£}34\,414.63}.
A1[1]
(b) Amount borrowed = 32\,000 - 2500 = 29\,500.
A1[1]
Solving for PMT with I\% = 6.4 gives −575.8206, so the instalment is \text{£}575.82.
A1[1]
Total repaid = 60 \times 575.82 = \textbf{\text{£}34\,549.24}.
A1[1]
(c) Option 1 has the lower total, by \textbf{\text{£}134.61}. M1A1
[2]
(d) Equal totals with the same term means the same instalment, £573.58.
M1[1]
Solve for PV: N = 60, I\% = 6.4, PMT = -573.5772, FV = 0, giving a bank loan of 29 385.0665.
A1[1]
Rebate = 32\,000 - 29\,385.07 = \textbf{\text{£}2615} (to the nearest pound).
A1[1]
(e) Option 1: £573.58 is within £600; Option 2: £575.82 is within £600.
M1[1]
She can afford both options.
A1[1]
(f) For example: whether the dealer's rate is fixed, early-repayment charges, or whether she values keeping cash in hand.
R1[1]

Final answer

  • (a) GBP 573.58; GBP 34 414.63
  • (b) GBP 575.82; GBP 34 549.24
  • (c) Option 1, by GBP 134.61
  • (d) GBP 2615
  • (e) both
  • (f) for example, whether the rate is fixed, or any fees

Question 22

[13 marks]
Dalia borrows $200 000 to buy a flat. The loan is repaid by equal monthly payments over 25 years. The interest rate is 4.8% per year, compounded monthly.
(a) Find Dalia's monthly payment.
[3]
(b) Find the total interest Dalia pays if she makes this payment for the full 25 years.
[2]
(c) Find the amount Dalia owes after 5 years of payments.
[2]
Dalia decides to pay $150 more each month than the payment in part (a), from the start.
(d) Find the number of months, to the nearest whole month, after which the loan is fully repaid (the last payment may be smaller than the others).
[3]
(e) Find the total interest Dalia pays with the larger payments, and the interest she saves compared with part (b).
[3]
Show solutionHide solution
(a) Solver entries: N=300, I\%=4.8, PV=200\,000, PMT=?, FV=0, P/Y=C/Y=12. M1A1
[2]
PMT = -1145.99, so the payment is \mathbf{\$1145.99} per month.
A1[1]
(b) Total repaid = 300 \times 1145.9939 = 343\,798.18 (using the unrounded payment).
M1[1]
Interest = 343\,798.18 - 200\,000 = \mathbf{\$143\,798.18}.
A1[1]
(c) Solver entries: N=60, I\%=4.8, PV=200\,000, PMT=-1145.99, FV=?, P/Y=C/Y=12.
M1[1]
FV = -176\,589.99, so she owes \mathbf{\$176\,589.99}.
A1[1]
(d) New payment = 1145.9939 + 150 = 1295.9939 \approx 1295.99 (the solver keeps the unrounded value). Solver entries: N=?, I\%=4.8, PV=200\,000, PMT=-1295.99, FV=0, P/Y=C/Y=12. M1A1
[2]
N = 240.60: so 240 full payments and a final, smaller, 241st payment; the loan is repaid after \mathbf{241} months.
A1[1]
(e) After 240 payments the balance is 771.62; one month later this has grown to 771.62 \times 1.004 = 774.71, which is the final payment.
M1[1]
Total paid = 240 \times 1295.9939 + 774.71 = 311\,813.25, so the interest is 311\,813.25 - 200\,000 = \mathbf{\$111\,813.25}.
A1[1]
Interest saved = 143\,798.18 - 111\,813.25 = \mathbf{\$31\,984.92}.
A1[1]

Final answer

  • (a) USD 1145.99
  • (b) USD 143 798.18
  • (c) USD 176 589.99
  • (d) 241 months
  • (e) USD 111 813.25; saves USD 31 984.92

Question 23

[9 marks]
Parents open an education account for their daughter that pays a nominal annual interest rate of 4.2%, compounded monthly. They pay in £400 at the end of every month for the first 8 years, and then increase this to £650 at the end of every month for the next 7 years. The account starts empty.
(a) Find the value of the account after the first 8 years.
[2]
(b) Find the value of the account after the full 15 years.
[3]
(c) Find the total amount paid in, and the total interest earned.
[2]
(d) Find the single fixed monthly payment, made for all 15 years, that would give the same final value.
[2]
Show solutionHide solution
(a)
N = 96, I\% = 4.2, PV = 0, PMT = -400, FV = ?, P/Y = C/Y = 12
Solving for FV gives 45 544.9565, so the value is \textbf{\text{£}45\,544.96}. M1A1
[2]
(b) For the second stage the opening balance becomes PV: N = 84, PV = -45\,544.9565, PMT = -650. M1A1
[2]
Solving for FV gives 124 426.4061, so the value is \textbf{\text{£}124\,426.41}.
A1[1]
(c) Paid in = 96 \times 400 + 84 \times 650 = \textbf{\text{£}93\,000}.
A1[1]
Interest = 124\,426.41 - 93\,000 = \textbf{\text{£}31\,426.41}.
A1[1]
(d) N = 180, PV = 0, FV = +124\,426.4061; solving for PMT gives −497.3950.
M1[1]
The fixed payment is \textbf{\text{£}497.39} per month.
A1[1]

Final answer

  • (a) GBP 45 544.96
  • (b) GBP 124 426.41
  • (c) GBP 93 000 paid in, GBP 31 426.41 interest
  • (d) GBP 497.39

Question 24

[13 marks]
Meera's daughter will start university in 8 years. The first-year fee is $18 400 now, and fees rise by 3.5% per year.
(a) Find the first-year fee in 8 years' time, to the nearest dollar.
[2]
Meera wants to have exactly this amount available in 8 years. She will pay equal amounts into an account at the end of every month. The account pays a nominal annual interest rate of 3%, compounded monthly.
(b) Find the monthly payment if the account starts with a balance of zero. Give your answer to the nearest cent.
[3]
Meera then finds that she can put $4000 into the account today as well.
(c) Find the value of the $4000 after 8 years.
[2]
(d) Find the new monthly payment, to the nearest cent, that Meera needs to make with this $4000 included.
[3]
(e) Find the total amount Meera pays into the account in part (d) (including the $4000), and the total interest earned.
[3]
Show solutionHide solution
(a) 18\,400(1.035)^8
M1[1]
= 24\,229.29 \approx \mathbf{\$24\,229}.
A1[1]
(b) Solver entries: N=96, I\%=3, PV=0, PMT=?, FV=24\,229.29, P/Y=C/Y=12. M1A1
[2]
PMT = -223.63 per month: \mathbf{\$223.63}.
A1[1]
(c) 4000\left(1 + \dfrac{0.03}{12}\right)^{96}
M1[1]
= \mathbf{\$5083.47} (solver: N=96, I\%=3, PV=-4000, PMT=0, FV=?, P/Y=C/Y=12).
A1[1]
(d) Solver entries: N=96, I\%=3, PV=-4000, PMT=?, FV=24\,229.29, P/Y=C/Y=12. M1A1
[2]
PMT = \mathbf{\$176.71} per month.
A1[1]
(e) Total paid in = 4000 + 96 \times 176.7077 = \mathbf{\$20\,963.93} (using the unrounded payment). M1A1
[2]
Interest = 24\,229.29 - 20\,963.93 = \mathbf{\$3265.35}.
A1[1]

Final answer

  • (a) USD 24 229
  • (b) USD 223.63
  • (c) USD 5083.47
  • (d) USD 176.71
  • (e) USD 20 963.93 paid in, USD 3265.35 interest

Question 25

[10 marks]
A strawberry farm picks 320 kg of strawberries on the first day of the season. The farm plans to increase the amount picked by the same number of kilograms each day for the 30 days of the season.
(a) The daily amount increases by 8 kg each day. Find the total amount picked in the 30 days.
[2]
The farm wants to pick at least 15 000 kg in the 30 days, with the daily amount increasing by a whole number of kilograms, k, each day.
(b) Find the least value of k.
[3]
(c) Using this value of k, find the first day on which more than 600 kg is picked.
[2]
(d) The packing shed can process at most 12 000 kg in the first 25 days. Determine whether the plan in part (b) stays within this limit.
[2]
(e) Comment on whether the model is realistic for the whole season.
[1]
Show solutionHide solution
(a) \frac{30}{2}(640+29\times8) M1 =13\,080 kg A1 (b) 15(640+29k)\ge15\,000 M1 k\ge12.41\ldots A1 k=12 gives 14 820 and k=13 gives 15 255, so k=13 A1 (c) 320+13(n-1)>600 M1 Day 22 gives 593 and day 23 gives 606, so day 23 A1 (d) S_{25}=\frac{25}{2}(640+24\times13)=11\,900 M1 This is at most 12 000, so the plan is within the limit A1 (e) For example, picking cannot keep rising all season, because weather and ripening limit it
R1

Final answer

  • (a) 13 080 kg
  • (b) k=13
  • (c) day 23
  • (d) 11 900 kg, within the limit
  • (e) picking cannot keep rising all season

Question 26

[18 marks]
A solar farm 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) Find the first year in which the output is less than 2300 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} M1 =2353.9\ldots, so 2350 MWh A1 (b) \frac{2600(1-0.991^{25})}{1-0.991} M1 =58\,441\ldots, so 58 400 MWh A1 (c) 2600(0.991)^{n-1}<2300 M1 Year 14 gives 2311.7 and year 15 gives 2290.9, so year 15 A1 (d) N=180, I\%=4.8, PV=2\,400\,000, FV=0, P/Y=C/Y=12 M1A1 PMT=-18\,729.95, so EUR 18 729.95 A1 (e) Income =92\times2600(0.991)^{n-1}=239\,200(0.991)^{n-1}, instalments 12\times18\,729.946\ldots=224\,759.36 M1 Year 7 income 226 570.36 is above 224 759.36; year 8 income 224 531.23 is below A1 Year 8 A1 (f) N=96, same I\%, PV and PMT, solve FV M1A1 EUR 1 334 030.06 A1 (g) 84\times18\,729.946\ldots-1\,334\,030.06 M1 = EUR 239 285.44 A1 (h) For example, panels may fail or be replaced, or the fall may not stay at 0.9%
R1

Final answer

  • (a) 2350 MWh
  • (b) 58 400 MWh
  • (c) year 15
  • (d) EUR 18 729.95
  • (e) year 8
  • (f) EUR 1 334 030.06
  • (g) EUR 239 285.44
  • (h) for example, panels may fail, or the fall may not stay at 0.9%

Going further

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

G1 An annuity is a finite geometric series

G1 Worked example

[3 marks]
Anjali pays USD 200 into a savings account at the end of every month for 3 years. The account adds interest of 0.4% at the end of each month.
(a) Show that the amount after 36 deposits is \dfrac{200(1.004^{36}-1)}{0.004}, by writing the account value as a geometric series.
[2]
(b) Check the value with the finance solver.
[1]
Show solutionHide solution
(a) The first deposit earns interest for 35 months and the last for none, so the amount is 200\left(1.004^{35}+1.004^{34}+\dots+1.004+1\right) M1 This is a geometric series with 36 terms, first term 1 and ratio 1.004, so the sum is 200\times\dfrac{1.004^{36}-1}{1.004-1}=\dfrac{200(1.004^{36}-1)}{0.004} A1AG (b) N=36, I\%=4.8, PV=0, PMT=-200, P/Y=C/Y=12 gives FV=7727.62, so the amount is USD 7727.62, the same as the formula
A1

Final answer

USD 7727.62

G2 Flat rates hide the cost

G2 Investigation

[17 marks]
(a) Rui pays EUR 150 into an account at the end of every month for 3 years. Interest of 0.3% is added at the end of each month. Show that the amount in the account immediately after the 36th payment is \dfrac{150(1.003^{36}-1)}{0.003}. [2] Evaluate this amount.
[1]
(b) Check your answer to (a) with the finance solver, stating the values you enter.
[1]
(c) A loan of EUR 10 000 is repaid by 48 equal payments P at the end of each month, with interest of 0.4% per month. Explain why 10\,000=\dfrac{P}{1.004}+\dfrac{P}{1.004^2}+\dots+\dfrac{P}{1.004^{48}}, and hence show that P=\dfrac{10\,000\times0.004}{1-1.004^{-48}}. [3] Find P.
[1]
(d) A shop offers the same EUR 10 000 at a “flat rate” of 5% a year for 4 years: 5% of EUR 10 000 is charged as interest for each of the 4 years, and the total is repaid in 48 equal monthly payments. Find the monthly payment.
[2]
(e) Find the nominal annual interest rate, compounded monthly, that this flat-rate loan actually charges.
[2]
(f) Use your answers to (c), (d) and (e) to explain why a flat rate makes a loan look cheaper than it is.
[2]
(g) Repeat (d) and (e) for a 2-year and a 6-year flat-rate loan at 5%. Describe how the actual rate compares with the flat rate.
[3]
Show solutionHide solution
(a) The first payment earns interest for 35 months and the last for none: 150\left(1.003^{35}+1.003^{34}+\dots+1\right) M1 A geometric series with 36 terms and ratio 1.003 has sum 150\times\dfrac{1.003^{36}-1}{1.003-1}=\dfrac{150(1.003^{36}-1)}{0.003} A1AG The value is EUR 5693.38
A1
(b) N=36, I\%=3.6, PV=0, PMT=-150, P/Y=C/Y=12 gives FV=5693.38, the same value
A1
(c) A payment made k months from now is worth \dfrac{P}{1.004^k} today, because each month multiplies a balance by 1.004 M1 The present values of the 48 payments add up to the amount borrowed, which gives the stated equation A1 The right-hand side is a geometric series with first term \dfrac{P}{1.004} and ratio \dfrac{1}{1.004}, so it equals \dfrac{P}{1.004}\times\dfrac{1-1.004^{-48}}{1-\frac{1}{1.004}}=\dfrac{P\left(1-1.004^{-48}\right)}{0.004}, and so P=\dfrac{10\,000\times0.004}{1-1.004^{-48}} A1AG P=229.39, so EUR 229.39
A1
(d) Interest =0.05\times10\,000\times4=2000, so the total repaid is EUR 12 000 M1 Monthly payment =\dfrac{12\,000}{48}= EUR 250
A1
(e) N=48, PV=10\,000, PMT=-250, FV=0, P/Y=C/Y=12, solve for I\% M1 I\%=9.24, so the loan charges 9.24% a year
A1
(f) The flat-rate payment of EUR 250 is higher than the EUR 229.39 paid on a 4.8% loan A1 The true rate of 9.24% is almost double the quoted 5%, because the flat rate charges interest on money that has already been repaid
A1
(g) 2 years: interest 0.05\times10\,000\times2=1000, payment \dfrac{11\,000}{24}=458.33; 6 years: interest 3000, payment \dfrac{13\,000}{72}=180.56 M1 Actual rates 9.32% for 2 years and 9.06% for 6 years A1 The actual rate is about 1.8 to 1.9 times the flat rate, and a little lower for longer terms
A1

Final answer

  • (a) EUR 5693.38
  • (b) N=36, I\%=3.6, PV=0, PMT=-150, P/Y=C/Y=12
  • (c) EUR 229.39
  • (d) EUR 250 a month
  • (e) 9.24%
  • (f) the flat-rate payment is higher and the real rate is almost double 5%
  • (g) EUR 458.33, 9.32%; EUR 180.56, 9.06%; about 1.8 to 1.9 times the flat rate

From the book

Photon Guidebook: Mathematics Applications and Interpretation SL 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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