DP Maths AA HL - Number and Algebra - Arithmetic and Geometric Sequences

DP Maths AA HL - Number and Algebra - Arithmetic and Geometric Sequences - Full notes

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DP Maths AA HL | Arithmetic and Geometric Sequences

Arithmetic and Geometric Sequences

This topic contains 2 notes:

1.1 Sequences and Series
1.4 Financial Applications

DP MATHEMATICS AA · TOPIC 1: NUMBER AND ALGEBRA

1.1

Sequences and Series

Full notes for HL · Syllabus SL 1.1, 1.2, 1.3, 1.8

Contents

1	Standard form and significant figures
2	Sequences, recurrence relations and sigma notation
3	Arithmetic sequences
4	Arithmetic series
5	Geometric sequences
6	Geometric series
7	Infinite geometric series
8	Applications and mixed problems
9	Sequences defined recursively
10	Arithmetic and geometric sequences together
11	Proofs with series
12	Applications in science and finance
13	Additional exam-style practice
14	Exam-style questions with markscheme
15	Common misconceptions and links
16	Check your understanding
17	Summary and topic map
18	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
write numbers in the form a × 10k and calculate with them, with and without a GDC
describe sequences using an nth-term formula or a recurrence relation
read, write and evaluate sums in sigma notation
derive and use un = u1 + (n − 1)d and the arithmetic series formulas
derive and use un = u1rn−1 and the geometric series formulas
decide when an infinite geometric series converges and find its sum
solve problems where n is unknown, using logarithms or technology
model real situations with arithmetic and geometric sequences

This booklet covers SL 1.1, 1.2, 1.3, 1.8, which is common to SL and HL.

Guiding questions

How can patterns in numbers be described algebraically? When does adding infinitely many numbers give a finite answer? How do we choose between an arithmetic and a geometric model?

How this topic is examined

Sequences and series appear in almost every examination session, on both papers. Paper 1 (no calculator) asks for exact answers: terms, sums, sums to infinity, and algebraic conditions such as 'find k'. Paper 2 (calculator) adds modelling problems and questions in which n must be found from an inequality.

Many questions combine an arithmetic and a geometric sequence, or ask you to show that a sequence is arithmetic or geometric. Write down the formula you are using before you substitute: the method mark is often awarded for the correct formula with values substituted.

When n must be an integer, give the integer answer and explain your rounding: 'n > 22.1, so n = 23'.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Standard form and significant figures

A compact way to write very large and very small numbers.

Think about it

The mass of the Earth is about 5 970 000 000 000 000 000 000 000 kg, and the mass of a hydrogen atom is about 0.000 000 000 000 000 000 000 000 001 67 kg. How could you write both so that they are easy to read, compare and calculate with?

A number is in standard form (scientific notation) when it is written as a × 10k, where 1 ≤ a < 10 and k ∈ ℤ. The power k tells you how many places the decimal point has moved: positive k for large numbers, negative k for small numbers.

So 5 970 000 000 000 000 000 000 000 = 5.97 × 1024 and 0.000 042 = 4.2 × 10−5. To convert, place the decimal point after the first non-zero digit and count how many places it has moved.

Calculating in standard form. To multiply, multiply the a values and add the powers of 10; to divide, divide the a values and subtract the powers. Then adjust so that 1 ≤ a < 10. To add or subtract, first write both numbers with the same power of 10.

Significant figures. Unless a question says otherwise, IB answers must be exact or correct to three significant figures (3 s.f.). Leading zeros are not significant: 0.000 427 3 has four significant figures and is 0.000 427 to 3 s.f. Keep full calculator accuracy in intermediate steps and round only the final answer.

WORKED EXAMPLE 1 Converting (basic)

Write (a) 307 000 000 and (b) 0.000 060 5 in standard form.

Solution

(a) Decimal point moves 8 places left: 3.07 × 108A1

(b) Decimal point moves 5 places right: 6.05 × 10−5A1

WORKED EXAMPLE 2 Multiplying and dividing without a calculator (standard)

Find (a) (3 × 105) × (8 × 10−2); (b) (2.4 × 107) ÷ (6 × 10−3). Give answers in standard form.

Solution

(a) (3 × 8) × 105 − 2 = 24 × 103M1

= 2.4 × 104A1

(b) (2.4 ÷ 6) × 107 − (−3) = 0.4 × 1010M1

= 4 × 109A1

WORKED EXAMPLE 3 Adding numbers in standard form (exam level)

Without a calculator, find 3.2 × 106 + 4.5 × 105, giving your answer in standard form.

Solution

Write with the same power: 3.2 × 106 + 0.45 × 106M1

= 3.65 × 106A1

You cannot add the a values directly unless the powers of 10 are equal.

Common mistake

Leaving 24 × 103 or 0.4 × 1010 as a final answer. In standard form the first factor must satisfy 1 ≤ a < 10.

GDC tip

Calculators display standard form as, for example, 5.97E24. Never copy this notation into your answer; write 5.97 × 1024.

Practice exercise 1

1. Write 0.000 307 in standard form.

2. Write 4.06 × 10−3 as an ordinary number.

3. Calculate (6 × 108) ÷ (1.5 × 103).

4. Calculate (5 × 10−4)2.

5. Calculate 7.1 × 10−2 − 4 × 10−3.

6. Round 0.004 527 6 to 3 significant figures and write it in standard form.

7. Light travels 3.00 × 108 m s−1. How far does it travel in one year (365 days)? Give your answer in standard form to 3 s.f.

2 Sequences, recurrence relations and sigma notation

A sequence is a list; a series is a sum.

A sequence is an ordered list of numbers, called terms: u1, u2, u3, ... . The nth term un is called the general term. A sequence is finite if it stops and infinite if it continues forever.

A sequence can be defined in two ways. A formula for the nth term (explicit formula), such as un = 3n + 1, lets you find any term directly: u50 = 151. A recurrence relation (recursive formula) gives each term from the previous one, such as un+1 = un + 3 with u1 = 4. A recurrence relation needs a starting value, and to find u50 you would have to generate the previous 49 terms.

A series is the sum of the terms of a sequence. Sn denotes the sum of the first n terms: Sn = u1 + u2 + ... + un. Note that un = Sn − Sn−1: the nth term is the difference between consecutive sums.

Sigma notation uses the capital Greek letter Σ (sigma) to write a series compactly. The index (here k) runs through the integers from the lower limit to the upper limit:

Useful properties: the sum of a constant c from k = 1 to n is cn; constants can be taken outside the sum; and sums can be split: Σ(ak + bk) = Σak + Σbk.

WORKED EXAMPLE 4 Generating terms (basic)

(a) Find the first four terms of un = n2 − 2n. (b) A sequence is defined by u1 = 5, un+1 = 2un − 3. Find u4.

Solution

(a) u1 = −1, u2 = 0, u3 = 3, u4 = 8A1

(b) u2 = 7, u3 = 11, u4 = 19A1

WORKED EXAMPLE 5 Evaluating sigma notation (standard)

Evaluate the sum from k = 3 to 6 of (2k − 1)2.

Solution

Terms: k = 3: 25; k = 4: 49; k = 5: 81; k = 6: 121M1

Sum = 276A1

WORKED EXAMPLE 6 Writing in sigma notation (standard)

Write 4 + 7 + 10 + ... + 61 in sigma notation.

Solution

nth term: 3k + 1 (k = 1 gives 4)M1

Last term: 3k + 1 = 61 gives k = 20M1

Sum from k = 1 to 20 of (3k + 1)A1

WORKED EXAMPLE 7 From the sum to the nth term (exam level)

The sum of the first n terms of a sequence is Sn = 2n2 + 5n. Find un and show that the sequence is arithmetic.

Solution

un = Sn − Sn−1 = 2n2 + 5n − [2(n − 1)2 + 5(n − 1)]M1

= 2n2 + 5n − 2n2 + 4n − 2 − 5n + 5 = 4n + 3A1

un+1 − un = 4, a constant, so the sequence is arithmetic with d = 4R1

Common mistake

Miscounting terms in sigma notation. The sum from k = 3 to k = 10 has 10 − 3 + 1 = 8 terms, not 7.

Practice exercise 2

1. Find u10 for un = (−1)n(n + 2).

2. A sequence has u1 = 2 and un+1 = un2 − 1. Find u4.

3. Evaluate the sum from k = 1 to 5 of k3.

4. How many terms are in the sum from r = 7 to 31?

5. Write 1/2 + 1/4 + 1/8 + ... + 1/1024 in sigma notation.

6. Given Sn = n2 + 4n, find u6.

7. Show that the sum from k = 1 to n of (3k + 2) equals 3 × (the sum from k = 1 to n of k) + 2n.

3 Arithmetic sequences

Add the same number each time.

In an arithmetic sequence (arithmetic progression) the difference between consecutive terms is constant. This common difference is d = un+1 − un. The sequence 5, 8, 11, 14, ... has u1 = 5 and d = 3. If d > 0 the terms increase; if d < 0 they decrease; if d = 0 the sequence is constant.

Starting from u1, each new term adds one more d: u2 = u1 + d, u3 = u1 + 2d, u4 = u1 + 3d. To reach the nth term we add d a total of (n − 1) times. So:

The nth term is a linear function of n: plotting un against n gives points on a straight line with gradient d. This is why arithmetic sequences model linear growth: a salary rising by the same amount each year, or seats increasing by the same number each row.

Three consecutive terms. a, b, c are consecutive terms of an arithmetic sequence if and only if b − a = c − b, that is 2b = a + c: the middle term is the arithmetic mean of its neighbours.

Figure 1. The terms of an arithmetic sequence lie on a straight line with gradient d = 3.

WORKED EXAMPLE 8 A term and the number of terms (basic)

An arithmetic sequence has u1 = 7 and d = 4. (a) Find u20. (b) Which term equals 199?

Solution

(a) u20 = 7 + 19 × 4 = 83A1

(b) 7 + (n − 1)4 = 199, so (n − 1)4 = 192M1

n − 1 = 48, so it is the 49th termA1

WORKED EXAMPLE 9 Using two given terms (standard)

In an arithmetic sequence, u5 = 17 and u12 = 45. Find u1 and d.

Solution

u1 + 4d = 17 and u1 + 11d = 45M1

Subtract: 7d = 28, so d = 4A1

u1 = 17 − 16 = 1A1

WORKED EXAMPLE 10 Consecutive terms (standard)

The terms 2k + 1, 5k − 2 and 4k + 5 are consecutive terms of an arithmetic sequence. Find k and the three terms.

Solution

2(5k − 2) = (2k + 1) + (4k + 5)M1

10k − 4 = 6k + 6, so k = 2.5A1

Terms: 6, 10.5, 15 (common difference 4.5)A1

WORKED EXAMPLE 11 How many terms satisfy a condition (exam level)

The sequence 200, 193, 186, ... is arithmetic. (a) Find the first negative term. (b) How many terms are between 100 and 150?

Solution

(a) un = 200 − 7(n − 1) = 207 − 7n < 0 gives n > 29.6, so n = 30: u30 = −3M1 A1

(b) 100 < 207 − 7n < 150 gives 8.14 < n < 15.3M1

n = 9, 10, ..., 15: 7 termsA1

Common mistake

Writing un = u1 + nd. The first term already counts as one term, so only (n − 1) differences are added.

Practice exercise 3

1. Find the 50th term of 3, 7, 11, ... .

2. Find d if u3 = 10 and u8 = 35.

3. Which term of 4, 11, 18, ... is 704?

4. The 7th term is 26 and the 15th term is 58. Find the 1st term.

5. Find x if x, 2x + 3 and 5x − 4 are consecutive terms of an arithmetic sequence.

6. Find the number of terms in 5, 9, 13, ..., 401.

7. Find the first term of 80, 76.5, 73, ... that is less than 0.

8. Show that un = 7 − 2n defines an arithmetic sequence and state d.

4 Arithmetic series

Pair the first and last terms.

The story goes that the young Carl Friedrich Gauss added 1 + 2 + ... + 100 in seconds by pairing terms: 1 + 100, 2 + 99, 3 + 98, ..., each pair totalling 101. There are 50 pairs, so the sum is 5050. The same idea works for any arithmetic series.

Why it works

Write Sn forwards and backwards: Sn = u1 + (u1 + d) + ... + un and Sn = un + (un − d) + ... + u1. Adding the two lines term by term, every pair totals u1 + un, and there are n pairs, so 2Sn = n(u1 + un). Substituting un = u1 + (n − 1)d gives the second form.

Use the first form when you know the last term, and the second when you know d. Sn is a quadratic in n with no constant term, so Sn = An2 + Bn for an arithmetic series; conversely, any sequence with a sum of this form is arithmetic.

WORKED EXAMPLE 12 Sum of the first n terms (basic)

Find the sum of the first 25 terms of 3, 7, 11, ... .

Solution

S25 = 25/2 × (2 × 3 + 24 × 4)M1

= 12.5 × 102 = 1275A1

WORKED EXAMPLE 13 Multiples between two values (standard)

Find the sum of all the multiples of 7 between 100 and 300.

Solution

First multiple 105, last 294; d = 7M1

Number of terms: 105 + (n − 1)7 = 294 gives n = 28A1

S28 = 28/2 × (105 + 294) = 5586A1

WORKED EXAMPLE 14 Finding n from a sum (standard)

The sum of the first n terms of 4, 9, 14, ... is 1030. Find n.

Solution

n/2 × (2 × 4 + 5(n − 1)) = 1030M1

n(5n + 3) = 2060, so 5n2 + 3n − 2060 = 0A1

n = 20 or n = −20.6; n must be a positive integer, so n = 20A1

WORKED EXAMPLE 15 Two unknowns from sums (exam level)

An arithmetic series has S5 = 40 and S10 = 155. Find u1 and d.

Solution

5/2 (2u1 + 4d) = 40 gives u1 + 2d = 8M1

10/2 (2u1 + 9d) = 155 gives 2u1 + 9d = 31M1

Solve: 5d = 15, d = 3, u1 = 2A1 A1

WORKED EXAMPLE 16 Smallest n for a sum to exceed a value (exam level, GDC)

For the series 2 + 5 + 8 + ..., find the least n for which Sn > 1000.

Solution

n/2 (4 + 3(n − 1)) > 1000, so 3n2 + n − 2000 > 0M1

Positive root n = 25.65A1

n = 26A1

GDC tip

To find the least n with Sn > 1000, either solve the quadratic with the polynomial solver, or use a table: enter y = x/2 (4 + 3(x − 1)) and scroll to find where y first exceeds 1000. A table also lets you check your answer.

Practice exercise 4

1. Find the sum of the first 40 positive integers.

2. Find S10 for 2, 5, 8, ... .

3. Find the sum of all odd numbers from 1 to 99.

4. Find the sum of the arithmetic series 50 + 47 + 44 + ... + 2.

5. How many terms of 6 + 10 + 14 + ... are needed for a sum of 510?

6. An arithmetic series has u1 = 12 and S20 = 620. Find d.

7. Find the least n for which the sum of 1 + 4 + 7 + ... exceeds 5000.

8. Prove that the sum of the first n odd numbers is n2.

5 Geometric sequences

Multiply by the same number each time.

In a geometric sequence (geometric progression) each term is the previous term multiplied by a constant, the common ratio r = un+1 ÷ un. The sequence 3, 6, 12, 24, ... has r = 2; 80, 40, 20, ... has r = ½; 2, −6, 18, −54, ... has r = −3, so the signs alternate.

Each term multiplies by one more factor of r: u2 = u1r, u3 = u1r2. To reach the nth term we multiply by r a total of (n − 1) times:

The behaviour depends on r: r > 1 gives exponential growth; 0 < r < 1 gives exponential decay towards 0; −1 < r < 0 gives terms that alternate in sign and shrink; r < −1 gives alternating terms that grow in size. Geometric sequences model percentage change: growing by 5% means multiplying by 1.05.

Three consecutive terms. a, b, c are consecutive terms of a geometric sequence if b/a = c/b, that is b2 = ac: b is the geometric mean of a and c.

Figure 2. A geometric sequence with r = 2 grows exponentially.

WORKED EXAMPLE 17 Finding a term (basic)

Find the 10th term of 5, 15, 45, ... .

Solution

r = 15 ÷ 5 = 3A1

u10 = 5 × 39 = 98 415A1

WORKED EXAMPLE 18 Using two terms (standard)

A geometric sequence has u3 = 18 and u6 = −486. Find r and u1.

Solution

u6 ÷ u3 = r3 = −27M1

r = −3A1

u1 = 18 ÷ (−3)2 = 2A1

WORKED EXAMPLE 19 Consecutive terms (standard)

The numbers k − 3, k and 3k + 4 are consecutive terms of a geometric sequence. Find the possible values of k.

Solution

k2 = (k − 3)(3k + 4)M1

k2 = 3k2 − 5k − 12, so 2k2 − 5k − 12 = 0A1

(2k + 3)(k − 4) = 0: k = 4 or k = −1.5A1

WORKED EXAMPLE 20 Finding n with logarithms (exam level)

In the geometric sequence 2, 6, 18, ..., which is the first term greater than 10 000?

Solution

2 × 3n−1 > 10 000, so 3n−1 > 5000M1

n − 1 > log 5000 ÷ log 3 = 7.75M1

n > 8.75, so the 9th term (13 122)A1

Common mistake

Using rn instead of rn−1, or forgetting that an even power of a negative ratio is positive.

Practice exercise 5

1. Find r and u6 for 64, 32, 16, ... .

2. A geometric sequence has u2 = 6 and u5 = 162. Find r and u1.

3. Find the 8th term of 3, −6, 12, ... .

4. Which term of 5, 10, 20, ... is 2560?

5. Find x > 0 if x, 6 and x + 9 are consecutive terms of a geometric sequence.

6. Find the first term of 1000, 900, 810, ... that is less than 100.

7. u4 = 24 and u7 = 3. Find u10.

6 Geometric series

A neat formula from subtracting rSn.

Why it works

Write Sn = u1 + u1r + u1r2 + ... + u1rn−1. Multiply by r: rSn = u1r + u1r2 + ... + u1rn. Subtracting, every term except the first and last cancels: Sn − rSn = u1 − u1rn, so Sn(1 − r) = u1(1 − rn). Dividing by (1 − r), valid for r ≠ 1, gives the formula.

Both forms are equivalent; use the first when r > 1 and the second when r < 1 to avoid negative numerators and denominators. If r = 1, every term equals u1 and Sn = nu1.

WORKED EXAMPLE 21 Sum of a geometric series (basic)

Find the sum of the first 8 terms of 3, 6, 12, ... .

Solution

S8 = 3(28 − 1) ÷ (2 − 1)M1

= 3 × 255 = 765A1

WORKED EXAMPLE 22 Sigma notation (standard)

Evaluate the sum from k = 1 to 10 of 5(0.8)k, giving your answer to 3 s.f.

Solution

First term u1 = 5 × 0.8 = 4, r = 0.8, n = 10M1

S10 = 4(1 − 0.810) ÷ 0.2M1

= 17.9A1

WORKED EXAMPLE 23 Smallest n for a sum (exam level)

Find the least number of terms of 1 + 1.5 + 2.25 + ... needed for the sum to exceed 1000.

Solution

(1.5n − 1) ÷ 0.5 > 1000, so 1.5n > 501M1

n > log 501 ÷ log 1.5 = 15.3A1

n = 16A1

WORKED EXAMPLE 24 Finding the ratio from sums (exam level)

For a geometric series with r ≠ 1, S6 = 9S3. Find r.

Solution

u1(r6 − 1)/(r − 1) = 9u1(r3 − 1)/(r − 1)M1

r6 − 1 = (r3 − 1)(r3 + 1), so r3 + 1 = 9M1

r3 = 8, r = 2A1

Practice exercise 6

1. Find S5 for 2, −4, 8, ... .

2. Find the sum of the first 12 terms of 81, 27, 9, ..., to 3 s.f.

3. Evaluate the sum from k = 0 to 7 of 3 × 2k.

4. How many terms of 4 + 12 + 36 + ... are needed for the sum to exceed 1 000 000?

5. A geometric series has u1 = 6 and S3 = 42. Find the possible values of r.

6. Explain why the formula cannot be used when r = 1, and state Sn in that case.

7 Infinite geometric series

Adding forever can give a finite total.

Think about it

Walk halfway to a wall, then half the remaining distance, then half again, forever. You take infinitely many steps. How far do you walk in total?

If |r| < 1, the terms of a geometric sequence get smaller and smaller, and rn → 0 as n → ∞. In the formula Sn = u1(1 − rn)/(1 − r), the rn term disappears, so Sn approaches a limit called the sum to infinity:

The series converges. If |r| ≥ 1, the terms do not shrink to zero and the sum grows without limit or oscillates: the series diverges and has no sum to infinity. In the walking example, 1/2 + 1/4 + 1/8 + ... = (1/2) ÷ (1 − 1/2) = 1: you reach the wall.

Recurring decimals are infinite geometric series: 0.777... = 0.7 + 0.07 + 0.007 + ... = 0.7 ÷ (1 − 0.1) = 7/9.

When r contains a variable (for example r = 2x − 3), the series converges only for the values of x satisfying |r| < 1, a condition for convergence that you must state and solve.

Figure 3. Partial sums approach the sum to infinity.

WORKED EXAMPLE 25 Sum to infinity (basic)

Find the sum to infinity of 12 + 4 + 4/3 + ... .

Solution

r = 1/3; |r| < 1 so the series convergesR1

S∞ = 12 ÷ (1 − 1/3) = 18A1

WORKED EXAMPLE 26 Recurring decimal (standard)

Write 0.272727... as a fraction in its simplest form.

Solution

0.27 + 0.0027 + ...: u1 = 0.27, r = 0.01M1

S∞ = 0.27 ÷ 0.99 = 27/99 = 3/11A1

WORKED EXAMPLE 27 Condition for convergence (standard)

For which values of x does 1 + (2x − 3) + (2x − 3)2 + ... converge? Find the sum when x = 1.75.

Solution

|2x − 3| < 1 gives −1 < 2x − 3 < 1M1

1 < x < 2A1

x = 1.75: r = 0.5, S∞ = 1 ÷ 0.5 = 2A1

WORKED EXAMPLE 28 Using the sum to infinity and another term (exam level)

A geometric series has S∞ = 27 and u2 = 6. Find the possible values of r and u1.

Solution

u1/(1 − r) = 27 and u1r = 6, so 6/r = 27(1 − r)M1

27r2 − 27r + 6 = 0, 9r2 − 9r + 2 = 0, (3r − 1)(3r − 2) = 0A1

r = 1/3, u1 = 18; or r = 2/3, u1 = 9 (both satisfy |r| < 1)A1 R1

Key point

Always state |r| < 1 when you use the sum to infinity, and check it for every solution.

Practice exercise 7

1. Find S∞ for 50 − 25 + 12.5 − ... .

2. Write 0.444... and 0.123123... as fractions.

3. For which x does 3 + 3x/2 + 3x2/4 + ... converge?

4. A geometric series has S∞ = 5 and u1 = 2. Find r.

5. S∞ = 3S3 for a series with |r| < 1. Show that r3 = 2/3.

6. Find the least n for which S∞ − Sn < 0.001 for 1 + 0.5 + 0.25 + ... .

8 Applications and mixed problems

Linear growth is arithmetic; percentage growth is geometric.

To choose a model, look at how the quantity changes. If it changes by a fixed amount each period (a salary rising by $2000 a year, a car park adding 15 cars per hour), it forms an arithmetic sequence. If it changes by a fixed percentage (a population growing 3% per year, a car losing 15% of its value per year), it forms a geometric sequence: increasing by p% means multiplying by 1 + p/100; decreasing by p% means multiplying by 1 − p/100.

Read the question carefully: is n the year number or the number of years after the start? Is the first term u1 or u0? A table of the first few values avoids off-by-one errors.

WORKED EXAMPLE 29 Bouncing ball (standard)

A ball is dropped from 10 m. After each bounce it rises to 70% of its previous height. Find the total distance travelled before it comes to rest.

Solution

Down: 10 + 7 + 4.9 + ... = 10 ÷ 0.3 = 33.3 mM1

Up: 7 + 4.9 + ... = 7 ÷ 0.3 = 23.3 mM1

Total = 56.7 m (exactly 170/3 m)A1

WORKED EXAMPLE 30 Comparing salary offers (exam level)

Job A: starting salary $40 000, rising by $2500 per year. Job B: starting salary $38 000, rising by 6% per year. (a) Find each salary in year 10. (b) Find the total earned over 10 years for each. (c) In which year does B's salary first exceed A's?

Solution

(a) A: 40 000 + 9 × 2500 = $62 500; B: 38 000 × 1.069 = $64 200.20A1 A1

(b) A: 10/2 (40 000 + 62 500) = $512 500; B: 38 000(1.0610 − 1)/0.06 = $500 870.21A1 A1

Job B's annual salary overtakes job A's from year 9, but job A's total over 10 years is still higher. Questions like this test interpretation as well as calculation.

Figure 4. Geometric growth eventually overtakes arithmetic growth.

WORKED EXAMPLE 31 Mixed arithmetic and geometric (exam level)

The first, second and fourth terms of an arithmetic sequence (d ≠ 0) form the first three terms of a geometric sequence. Given u1 = 3, find d and the common ratio.

Solution

Arithmetic terms: 3, 3 + d, 3 + 3dM1

Geometric: (3 + d)2 = 3(3 + 3d), so d2 + 6d + 9 = 9 + 9dM1

d2 − 3d = 0, d ≠ 0: d = 3; terms 3, 6, 12: r = 2A1 A1

Practice exercise 8

1. A car worth $24 000 loses 15% of its value each year. Find its value after 5 years.

2. Seats: 20 in row 1, increasing by 3 per row. Find the total in 15 rows.

3. A population of 2000 grows by 4% per year. In which year does it first exceed 3000?

4. A runner trains 5 km in week 1 and increases by 1.5 km each week. In which week does she first run more than 30 km, and what is her total distance up to and including that week?

5. A pendulum swings 40 cm, and each swing is 90% of the previous one. Find the total distance swung before stopping.

6. The 2nd and 3rd terms of a geometric sequence with first term 4 are the 3rd and 7th terms of an arithmetic sequence with first term 4. Find r (r ≠ 1) and d.

9 Sequences defined recursively

Generating terms step by step.

A recursive definition gives a starting value and a rule linking each term to the previous one, such as u1 = 3, un+1 = 2un − 1. Arithmetic sequences have un+1 = un + d and geometric sequences un+1 = run. Other recursive rules generate sequences that are neither, and some (such as un+1 = 0.5un + 6) approach a limit.

A limit L of a convergent recursive sequence satisfies L = f(L), because when n is large, un+1 and un are both close to L.

WORKED EXAMPLE 32 Generating terms (basic)

u1 = 3, un+1 = 2un − 1. Find u2, u3, u4 and an explicit formula.

Solution

5, 9, 17A1

un − 1 = 2n: un = 2n + 1M1 A1

WORKED EXAMPLE 33 A limit (exam level)

u1 = 20, un+1 = 0.5un + 6. Find the limit of the sequence.

Solution

L = 0.5L + 6M1

L = 12A1

Practice exercise 9

1. u1 = 1, un+1 = 3un + 2. Find u4.

2. u1 = 100, un+1 = 0.8un + 10. Find the limit.

3. Show that un = 3 × 2n − 1 satisfies un+1 = 2un + 1.

10 Arithmetic and geometric sequences together

Problems linking the two.

Some questions involve terms that are simultaneously in arithmetic and geometric sequences, or two sequences sharing terms. Write each condition algebraically: for arithmetic, 2b = a + c; for geometric, b2 = ac; then solve the resulting equations.

WORKED EXAMPLE 34 Shared terms (exam level)

The first, third and seventh terms of an arithmetic sequence (d ≠ 0) form a geometric sequence. Find the common ratio.

Solution

(a + 2d)2 = a(a + 6d)M1

4ad + 4d2 = 6ad, so a = 2dA1

Terms 2d, 4d, 8d: r = 2A1

WORKED EXAMPLE 35 Comparing growth (standard)

Sequence A: 100, 110, 120, ...; sequence B: 10, 12, 14.4, .... Find the first n with B's nth term greater than A's.

Solution

10(1.2)n−1 > 100 + 10(n − 1)M1

Table: n = 19: 266.2 vs 280; n = 20: 319.5 vs 290A1

n = 20A1

Practice exercise 10

1. x, x + 6 and 4x are consecutive terms of a geometric sequence (x > 0). Find x.

2. The 2nd, 5th and 14th terms of an arithmetic sequence form a geometric sequence. Find the ratio.

11 Proofs with series

General results.

Results about series can be proved algebraically: write Sn in terms of u1, d (or r) and n, and manipulate. Many identities follow from the relationship un = Sn − Sn−1.

WORKED EXAMPLE 36 Sum of odd numbers (standard)

Prove that the sum of the first n odd numbers is n2.

Solution

u1 = 1, d = 2A1

Sn = n/2 (2 + 2(n − 1)) = n2M1 AG

WORKED EXAMPLE 37 Geometric identity (exam level)

Show that for a geometric series, S2n/Sn = 1 + rn.

Solution

S2n = u1(r2n − 1)/(r − 1)A1

r2n − 1 = (rn − 1)(rn + 1)M1

So S2n/Sn = rn + 1AG

Practice exercise 11

1. Prove that the sum of the first n even numbers is n(n + 1).

2. Show that a geometric series with r = −1 and an even number of terms has sum 0.

12 Applications in science and finance

Modelling with sequences.

Sequences model quantities measured at regular intervals: drug levels after repeated doses, populations, depreciation, and loan balances. When a fixed amount is added and a fixed proportion removed each period, the recursive rule un+1 = run + c applies, and the long-term level is c/(1 − r).

WORKED EXAMPLE 38 Repeated doses (exam level)

A patient takes 100 mg of a drug every 12 hours; 40% of the drug in the body remains after 12 hours. Find the amount just after the 5th dose and the long-term amount.

Solution

un = 100(1 + 0.4 + ... + 0.4n−1)M1

u5 = 100(1 − 0.45)/0.6 = 165 mgA1

Long term = 100/0.6 = 167 mgA1

Practice exercise 12

1. A lake contains 5000 fish. Each year 20% are caught and 600 are added. Find the long-term population.

2. A ball dropped from 3 m rebounds to 60% of its height. Find the total distance travelled before it stops.

13 Additional exam-style practice

Mixed Paper 1 and Paper 2 questions.

WORKED EXAMPLE 39 Paper 1 question (exam level)

The sum of the first n terms of a sequence is Sn = 2n2 + 3n. (a) Find u1 and u2. (b) Show that the sequence is arithmetic.

Solution

(a) u1 = 5, u2 = S2 − S1 = 14 − 5 = 9A1 A1

(b) un = Sn − Sn−1 = 4n + 1, linear in nM1 R1

WORKED EXAMPLE 40 Paper 2 question (exam level)

A geometric series has first term 64 and sum to infinity 256. (a) Find r. (b) Find the least n for which Sn > 250.

Solution

(a) 64/(1 − r) = 256: r = 0.75A1

(b) 256(1 − 0.75n) > 250: 0.75n < 0.0234M1

n > 13.06: n = 14A1

Practice exercise 13

1. Find the sum of all integers between 1 and 500 that are divisible by 3 or 5.

2. The first three terms of a geometric sequence are k, k + 4, 2k + 8 with k ≠ −4. Find k.

14 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style, no calculator) (about 18 minutes)

The first three terms of a geometric sequence are 4x, x + 3 and x, where x ≠ 0.

(a) Show that 3x2 − 6x − 9 = 0. [2]

(b) Hence find the two possible values of x. [2]

(d) For the negative value of x, find the sum of the first 5 terms. [3]

Markscheme

Part	Answer	Marks
(a)	(x + 3)2 = 4x × x	M1
	x2 + 6x + 9 = 4x2, so 3x2 − 6x − 9 = 0	AG
(b)	3(x − 3)(x + 1) = 0	M1
	x = 3 or x = −1	A1
(c)	terms 12, 6, 3: r = 1/2	A1
	S∞ = 12 ÷ (1 − 1/2)	M1
	= 24	A1
(d)	terms −4, 2, −1: r = −1/2	A1
	S5 = −4(1 − (−1/2)5) ÷ (3/2)	M1
	= −11/4	A1

Examiner's comments. In (a), the answer is given, so every algebraic step must be shown. In (d), sign errors with the negative ratio were common.

Question 2 (Paper 2 style, calculator) (about 15 minutes)

Sam saves $50 in month 1 and increases the amount he saves by $8 each month. Ana saves $50 in month 1 and increases the amount she saves by 6% each month.

(a) Find the amount each saves in month 12. [3]

(b) Find the total each has saved after 12 months. [3]

(d) Find the first month at the end of which Ana's total savings exceed Sam's. [2]

Markscheme

Part	Answer	Marks
(a)	Sam: 50 + 11 × 8 = $138	A1
	Ana: 50 × 1.0611	M1
	= $94.91	A1
(b)	Sam: 12/2 (50 + 138) = $1128	A1
	Ana: 50(1.0612 − 1)/0.06	M1
	= $843.50	A1
(c)	50 × 1.06n−1 > 50 + 8(n − 1): by table or solver	M1
	month 32	A1
(d)	compare Sn for both sequences	M1
	month 43	A1

Examiner's comments. Parts (c) and (d) are best done with a GDC table. Candidates who confused the monthly amount with the running total lost marks in (d).

15 Common misconceptions and links

Students often think…	Actually…
un = u1 + nd	un = u1 + (n − 1)d
every geometric series has a sum to infinity	only when \|r\| < 1
a 5% increase means r = 0.05	r = 1.05
n can be any real number	n must be a positive integer
Sn and un are the same	un = Sn − Sn−1

How 1.1 connects to the rest of the course

Topic	Connection
1.4 Financial applications	compound interest is geometric
1.5 Exponents and logarithms	solving for n
1.15 Proof by induction (HL)	proving sum formulas
5.19 Maclaurin series (HL)	infinite series

16 Check your understanding

Section A: multiple choice

1. The 10th term of 2, 5, 8, ... is

A. 29 B. 32 C. 27 D. 30

2. The common ratio of 9, −3, 1, ... is

A. 1/3 B. −3 C. −1/3 D. 3

3. S∞ of 6 + 3 + 1.5 + ... is

A. 10.5 B. 12 C. 9 D. no sum

4. The sum from k = 1 to 5 of k2 is

A. 25 B. 30 C. 15 D. 55

5. A geometric series converges if

A. r > 1 B. |r| < 1 C. r = 1 D. |r| ≥ 1

6. 4.5 × 10−3 as a decimal is

A. 0.0045 B. 0.045 C. 4500 D. 0.00045

Section B: structured

7. Prove the formula for the sum of a geometric series. [3]

8. A population of 2000 grows by 4% per year. In which year does it first exceed 3000? [3]

9. Find the sum of the arithmetic series 50 + 47 + 44 + ... + 2. [3]

Answers

1. A 2. C 3. B 4. D 5. B 6. A

7. Write Sn and rSn; subtract; Sn(1 − r) = u1(1 − rn); divide.

8. 2000 × 1.04n > 3000; n > 10.3; during the 11th year.

9. n = 17; S = 17/2 × 52 = 442.

17 Summary and topic map

Idea	Remember
Standard form	a × 10k, 1 ≤ a < 10
Sigma	Σ from lower to upper; terms = upper − lower + 1
Arithmetic	un = u1 + (n − 1)d; Sn = n/2 (u1 + un)
Geometric	un = u1rn−1; Sn = u1(rn − 1)/(r − 1)
Sum to infinity	u1/(1 − r), only if \|r\| < 1
Consecutive terms	arithmetic 2b = a + c; geometric b2 = ac
Applications	fixed amount: arithmetic; fixed percentage: geometric

Before the exam, make sure you…

•write the formula before substituting
•check n is a positive integer and round correctly
•state |r| < 1 for sums to infinity
•convert percentage changes to multipliers
•use a GDC table for 'first n' questions

Sequences and Series on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 5. Topic map for Sequences and Series.

18 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. 3.07 × 10−4

2. 0.004 06

3. 4 × 105

4. 2.5 × 10−7

5. 6.7 × 10−2

6. 4.53 × 10−3

7. 9.46 × 1015 m

Practice exercise 2

1. 12

2. u2 = 3, u3 = 8, u4 = 63

3. 225

4. 25

5. sum from k = 1 to 10 of (1/2)k

6. S6 − S5 = 60 − 45 = 15

7. split the sum and take constants out: 3Σk + Σ2 = 3Σk + 2n

Practice exercise 3

1. 199

2. 5

3. 101st

4. 2

5. x = 5

6. 100

7. u24 = −0.5

8. un+1 − un = −2, constant; d = −2

Practice exercise 4

1. 820

2. 155

3. 2500

4. n = 17; S = 442

5. n = 15

6. d = 2

7. n = 58

8. Sn = n/2 (2 + 2(n − 1)) = n2

Practice exercise 5

1. r = ½, u6 = 2

2. r = 3, u1 = 2

3. −384

4. 10th term

5. x = 3

6. u23 ≈ 98.5

7. 0.375

Practice exercise 6

1. 22

2. 121

3. 765

4. n = 12

5. r = 2 or r = −3

6. denominator r − 1 = 0; Sn = nu1

Practice exercise 7

1. 100/3

2. 4/9; 41/333

3. −2 < x < 2

4. 0.6

5. u1/(1 − r) = 3u1(1 − r3)/(1 − r) gives 1 = 3(1 − r3)

6. 2 × 0.5n < 0.001 gives n = 11

Practice exercise 8

1. $10 649 (to the nearest dollar)

2. 615

3. 4% growth: n > 10.3, so during the 11th year

4. week 18 (30.5 km); total 319.5 km

5. 400 cm

6. 4 + 2d = 4r and 4 + 6d = 4r2 give r2 − 3r + 2 = 0: r = 2, d = 2

Practice exercise 9

1. 53

2. 50

3. 2(3 × 2n − 1) + 1 = 3 × 2n+1 − 1

Practice exercise 10

1. (x + 6)2 = 4x2: x = 6

2. r = 3

Practice exercise 11

1. Sn = n/2 (4 + 2(n − 1)) = n(n + 1)

2. (−1)n = 1 for even n, so Sn = u1(1 − 1)/2 = 0

Practice exercise 12

1. 600/0.2 = 3000

2. 3 + 2(1.8)/(0.4) = 12 m

Practice exercise 13

1. 41 583 + 24 750 − 8415 = 57 918 (multiples of 3: 166 terms; of 5: 99 terms; of 15: 33 terms)

2. (k + 4)2 = 2k(k + 4), so k + 4 = 2k: k = 4

DP MATHEMATICS AA · TOPIC 1: NUMBER AND ALGEBRA

1.4

Financial Applications

Full notes for HL · Syllabus SL 1.4

Contents

1	Simple and compound growth
2	Compound interest
3	Compounding periods
4	Finding the time, rate or principal
5	Depreciation
6	Inflation and real value
7	Using the financial solver (TVM)
8	Loans and annuities
9	Compound interest as a geometric sequence
10	Compounding frequency and nominal rates
11	Effective annual rate
12	Depreciation
13	Inflation and real value
14	Loans and repayments
15	Mortgages and the effect of the term
16	Savings plans and annuities
17	Finding the number of payments or the rate
18	Investment decisions
19	Mixed review: worked problems
20	Additional exam-style practice
21	Exam-style questions with markscheme
22	Common misconceptions and links
23	Check your understanding
24	Summary and topic map
25	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
distinguish simple and compound growth and link them to arithmetic and geometric sequences
calculate compound interest with annual and more frequent compounding
find the time, rate or principal needed to reach a target value
calculate depreciation using a geometric model
adjust for inflation to find real values
use the financial (TVM) solver correctly, including the sign convention
calculate repayments and totals for loans and regular savings

This booklet covers SL 1.4, which is common to SL and HL.

Guiding questions

How can sequences model money growing or losing value over time? Why does the frequency of compounding matter? How do banks calculate loan repayments?

How this topic is examined

Financial applications are usually tested on Paper 2, where a GDC with a TVM solver is available. Questions ask for final values, interest rates, the number of years needed, comparisons between investment options, and loan repayments.

Show the formula or list every TVM input (N, I%, PV, PMT, FV, P/Y, C/Y) you used: without this, a wrong answer earns no method marks. Give money answers to two decimal places.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Simple and compound growth

Interest on interest.

Think about it

You invest $1000 at 5% per year. Will you have the same amount after 10 years whether interest is added to the balance or paid out to you each year?

With simple interest, interest is calculated on the original amount (the principal) only. $1000 at 5% earns $50 every year, so the balance after n years is 1000 + 50n: an arithmetic sequence. After 10 years it is $1500.

With compound interest, interest is added to the balance at the end of each period and then earns interest itself. The balance is multiplied by 1.05 each year: a geometric sequence. After 10 years it is 1000 × 1.0510 = $1628.89. The extra $128.89 is 'interest on interest'.

Over long periods, compound growth outstrips simple growth dramatically: after 40 years at 5%, simple interest gives $3000 while compound interest gives $7039.99.

Figure 1. Compound growth overtakes simple growth.

Practice exercise 1

1. Find the simple interest on $2400 at 3.5% per year for 6 years.

2. Find the value of $2400 at 3.5% compound interest per year after 6 years.

3. Which sequence type models a balance with simple interest, and why?

2 Compound interest

A geometric sequence of balances.

If a present value PV is invested at an annual interest rate of r% compounded once a year, the future value after n years is:

This is the nth term of a geometric sequence starting at PV (with u0 = PV) and ratio 1 + r/100. The interest earned is FV − PV.

WORKED EXAMPLE 1 Future value (basic)

Find the value of $4500 invested for 8 years at 3.2% per annum compounded annually, and the interest earned.

Solution

FV = 4500 × 1.0328M1

= $5789.62A1

Interest = 5789.62 − 4500 = $1289.62A1

WORKED EXAMPLE 2 Present value (standard)

How much must be invested now at 4% compounded annually to have $10 000 in 7 years?

Solution

PV × 1.047 = 10 000M1

PV = 10 000 ÷ 1.047M1

= $7599.18A1

Practice exercise 2

1. Find the value of $600 after 3 years at 10% compound interest.

2. Find the value of $15 000 after 12 years at 2.75% compounded annually.

3. How much must be invested now at 6% to have $5000 in 10 years?

3 Compounding periods

More frequent compounding, slightly more interest.

Interest is often compounded more than once a year: half-yearly (k = 2), quarterly (k = 4), monthly (k = 12) or daily (k = 365). The annual rate r% is divided equally between periods, so the rate per period is r/k %, and the number of periods is kn:

WORKED EXAMPLE 3 Monthly compounding (basic)

Find the value of $5000 after 6 years at 4% per annum compounded monthly.

Solution

FV = 5000 × (1 + 4/1200)72M1

= $6353.71A1

Compounding	k	Value of $1000 after 1 year at 12%
annually	1	$1120.00
half-yearly	2	$1123.60
quarterly	4	$1125.51
monthly	12	$1126.83
daily	365	$1127.47

The more frequent the compounding, the larger the final value, but the gains become very small. As k → ∞ (continuous compounding), the factor approaches er/100: $1000 at 12% compounded continuously gives 1000e0.12 = $1127.50.

Effective annual rate. To compare offers, compute the equivalent annual rate: (1 + r/(100k))k − 1. 12% compounded monthly is equivalent to 12.68% per year.

WORKED EXAMPLE 4 Comparing offers (exam level)

Bank A offers 5.1% compounded annually; bank B offers 5% compounded monthly. Which gives more after one year?

Solution

A: 1.051 per yearA1

B: (1 + 5/1200)12 = 1.0512M1 A1

Bank B is slightly better (5.12% effective)R1

Common mistake

Using r instead of r/k for each period, or n instead of kn for the number of periods.

Practice exercise 3

1. Find the value of $3000 after 2 years at 6% compounded quarterly.

2. Find the effective annual rate for 8% compounded quarterly.

3. $2000 is invested at 3% compounded half-yearly. Find its value after 5 years.

4. Which is better: 6% compounded annually or 5.85% compounded monthly?

4 Finding the time, rate or principal

Solve with logarithms or technology.

If the unknown is n, use logarithms (1.5) or the TVM solver. If the unknown is the rate, take the nth root. If the unknown is the principal, divide.

WORKED EXAMPLE 5 Finding the time (standard)

How many complete years are needed for $2000 to grow to more than $3000 at 4.5% per annum compounded annually?

Solution

2000 × 1.045n > 3000, so 1.045n > 1.5M1

n > ln 1.5 ÷ ln 1.045 = 9.21A1

10 yearsA1

WORKED EXAMPLE 6 Finding the rate (standard)

An investment of $8000 grows to $10 000 in 5 years with annual compounding. Find the interest rate.

Solution

8000(1 + r/100)5 = 10 000M1

1 + r/100 = 1.251/5 = 1.0456A1

r = 4.56% (3 s.f.)A1

WORKED EXAMPLE 7 Time with monthly compounding (exam level)

How many months will it take $5000 to reach $7000 at 3.6% per annum compounded monthly?

Solution

5000(1.003)m = 7000M1

m = ln 1.4 ÷ ln 1.003 = 112.3A1

113 monthsA1

Practice exercise 4

1. Find n if 1.06n = 2 (to 3 s.f.).

2. How long does it take money to double at 8% compounded annually?

3. Find the annual rate that turns $2500 into $4000 in 9 years.

4. How many quarters for $1000 to exceed $1500 at 5% compounded quarterly?

5 Depreciation

Losing value by a fixed percentage.

Depreciation is the loss in value of an asset (such as a car or machinery) over time. If the value decreases by r% each year, the value after n years is a geometric sequence with ratio 1 − r/100:

WORKED EXAMPLE 8 Depreciation (standard)

A machine bought for $45 000 depreciates by 18% per year. (a) Find its value after 4 years. (b) After how many complete years will its value first fall below $10 000?

Solution

(a) 45 000 × 0.824 = $20 345.48A1

(b) 45 000 × 0.82n < 10 000, n > ln(2/9) ÷ ln 0.82 = 7.58M1

8 yearsA1

WORKED EXAMPLE 9 Finding a depreciation rate (exam level)

A car bought for $32 000 is worth $14 000 after 6 years. Find the annual rate of depreciation.

Solution

32 000(1 − r/100)6 = 14 000M1

1 − r/100 = (14/32)1/6 = 0.8711A1

r = 12.9%A1

Practice exercise 5

1. A phone worth $1200 depreciates by 30% per year. Find its value after 3 years.

2. A boat loses 8% of its value per year. After how many years is it worth less than half its original value?

3. A $20 000 car is worth $9000 after 5 years. Find the rate of depreciation.

6 Inflation and real value

Money loses purchasing power.

Inflation is the general increase in prices over time. It reduces the purchasing power of money: if prices rise 3% per year, $100 buys less each year.

The real value of an investment measures what it can buy in today's prices. A simple approach used in the AA course: the real interest rate ≈ nominal interest rate − inflation rate. If savings earn 5% and inflation is 2%, real growth is about 3% per year. (A more precise version divides the growth factors: 1.05/1.02 = 1.0294, a real rate of 2.94%.)

To find what amount will be needed in future to buy the same goods, grow today's price at the inflation rate.

WORKED EXAMPLE 10 Real value (standard)

$10 000 is invested for 5 years at 4.5% per annum compounded annually. Inflation is 2.5% per year. Estimate the real value of the investment in today's money.

Solution

Real rate ≈ 4.5 − 2.5 = 2%M1

Real value ≈ 10 000 × 1.025 = $11 040.81A1

WORKED EXAMPLE 11 Future cost (standard)

A car costs $25 000 today. If inflation averages 3.2% per year, estimate its cost in 8 years.

Solution

25 000 × 1.0328M1

= $32 164.56A1

Practice exercise 6

1. Nominal rate 6%, inflation 3.5%. Approximate real rate?

2. $5000 is invested at 3% for 10 years with inflation 2%. Estimate its real value.

3. A coffee costs $4.50 today. Estimate its price in 15 years with 2.8% inflation.

7 Using the financial solver (TVM)

Let the GDC do the arithmetic, but enter it correctly.

On Paper 2, the time value of money (TVM) solver can find any one unknown. Inputs: N = total number of compounding periods (years × C/Y); I% = annual interest rate; PV = present value; PMT = regular payment (0 if none); FV = future value; P/Y = payments per year; C/Y = compounding periods per year.

Sign convention. Money leaving you (invested, deposited, repaid) is negative; money coming to you (received, withdrawn, borrowed) is positive. So an investment of $5000 is entered as PV = −5000 and FV comes out positive.

Input	Meaning	Example ($5000, 6 years, 4%, monthly)
N	number of periods	72
I%	annual rate	4
PV	present value (negative if invested)	−5000
PMT	regular payment	0
FV	future value	solve: 6353.71
P/Y, C/Y	periods per year	12, 12

GDC tip

If the solver gives an error, check the signs: PV and FV must have opposite signs. Also check that N is in periods, not years, and that P/Y and C/Y are set correctly before solving.

Common mistake

Entering PV and FV with the same sign, which gives an error, or entering N in years when compounding is monthly.

8 Loans and annuities

Regular payments.

A loan is repaid with regular payments that cover both interest and principal: this is amortization. The TVM solver finds the regular payment (PMT). For a loan, PV is positive (you receive the money), PMT is negative (you pay) and FV = 0 (the loan is fully repaid).

An annuity is a series of regular payments into or out of an account, for example regular savings or a pension. For regular savings, PMT is negative and FV is the accumulated value.

Total amount paid = PMT × N; total interest = total paid − amount borrowed.

WORKED EXAMPLE 12 Loan repayments (standard)

A $20 000 loan is repaid monthly over 5 years at 6% per annum compounded monthly. Find the monthly repayment and the total interest.

Solution

N = 60, I% = 6, PV = 20 000, FV = 0, P/Y = C/Y = 12M1

PMT = −386.66: $386.66 per monthA1

Total paid 60 × 386.66 = $23 199.60; interest $3199.60A1

WORKED EXAMPLE 13 Regular savings (standard)

Ana deposits $200 at the end of every month into an account paying 4.8% per annum compounded monthly. Find the value after 10 years.

Solution

N = 120, I% = 4.8, PV = 0, PMT = −200, P/Y = C/Y = 12M1

FV = $30 726.39A1

Total deposited $24 000; interest earned $6726.39A1

WORKED EXAMPLE 14 Loan term (exam level)

A $250 000 mortgage at 5.2% per annum compounded monthly is repaid at $1800 per month. How many years will it take?

Solution

I% = 5.2, PV = 250 000, PMT = −1800, FV = 0, P/Y = C/Y = 12M1

N = 213.0 monthsA1

17.7 years (213 payments)A1

Practice exercise 7

1. Find the monthly repayment on a $15 000 loan over 4 years at 7.5% compounded monthly.

2. How much interest is paid in total on that loan?

3. Find the value after 20 years of saving $150 per month at 5% compounded monthly.

4. What monthly payment is needed to repay $8000 in 2 years at 9% compounded monthly?

9 Compound interest as a geometric sequence

Linking finance to sequences.

With annual compounding at r%, the balances at the end of each year form a geometric sequence with first term PV(1 + r/100) and ratio 1 + r/100. This explains the formula FV = PV(1 + r/100)n, and allows questions about 'which year' to be answered with sequence techniques or tables.

WORKED EXAMPLE 15 Year of reaching a target (standard)

$4000 is invested at 3.8% compounded annually. In which year does the balance first exceed $6000?

Solution

4000(1.038)n > 6000M1

n > ln 1.5/ln 1.038 = 10.87A1

During year 11A1

WORKED EXAMPLE 16 Comparing simple and compound (standard)

Compare $5000 at 6% simple interest with $5000 at 5% compound interest after 10 and 20 years.

Solution

Simple: 8000 and 11 000A1

Compound: 8144.47 and 13 266.49A1

Compound overtakes simple before year 10R1

Practice exercise 8

1. Find the value of $2500 after 12 years at 4.5% compounded annually.

2. How many years does it take for money to triple at 7% compounded annually?

3. Find the rate needed for $1000 to grow to $1500 in 8 years with annual compounding.

10 Compounding frequency and nominal rates

Monthly, quarterly, daily.

A nominal annual rate of r% compounded k times per year gives r/k % per period and kn periods in n years. The more frequent the compounding, the larger the final value, but the gains diminish quickly.

Compounding	k	Value of $1000 after 1 year at 6%
annually	1	$1060.00
quarterly	4	$1061.36
monthly	12	$1061.68
daily	365	$1061.83

WORKED EXAMPLE 17 Quarterly compounding (standard)

Find the value of $7500 after 6 years at 4.8% compounded quarterly.

Solution

7500(1 + 0.048/4)24M1

= $9986.05A1

Practice exercise 9

1. Find the value of $3000 after 5 years at 3.6% compounded monthly.

2. Find the interest earned on $10 000 at 2.4% compounded quarterly for 3 years.

11 Effective annual rate

Comparing offers fairly.

The effective annual rate is the annual compound rate that gives the same growth as the nominal rate with its compounding. Banks quote it (often as the AER or APY) so that savers can compare products.

WORKED EXAMPLE 18 Comparing two accounts (exam level)

Account A pays 4.6% compounded annually; account B pays 4.5% compounded monthly. Which is better?

Solution

A: 4.6%; B: (1 + 0.045/12)12 − 1 = 4.594%M1 A1

A is very slightly betterR1

Practice exercise 10

1. Find the effective rate of 7% compounded quarterly.

2. Find the nominal rate compounded monthly that gives an effective rate of 5%.

12 Depreciation

Loss of value as a percentage.

Assets such as cars and machinery lose value over time. Reducing-balance depreciation at r% per year gives V = V0(1 − r/100)n, a geometric decrease. Straight-line depreciation (a fixed amount each year) is arithmetic.

WORKED EXAMPLE 19 Depreciation (standard)

A machine bought for $64 000 depreciates by 12% per year. Find its value after 6 years and when it first falls below $20 000.

Solution

64 000(0.88)6 = $29 721.86A1

0.88n < 0.3125: n > 9.10M1

After 10 yearsA1

WORKED EXAMPLE 20 Finding the rate (standard)

A car bought for $28 000 is worth $15 400 after 4 years. Find the annual depreciation rate.

Solution

(1 − r/100)4 = 0.55M1

r = 13.9%A1

Practice exercise 11

1. A laptop costs $1600 and depreciates by 25% per year. Find its value after 3 years.

2. Compare 20% reducing-balance depreciation with straight-line depreciation of $3000 per year for a $15 000 van after 4 years.

13 Inflation and real value

What will money buy?

Inflation reduces purchasing power. To find the real value of a future amount in today's money, divide by (1 + i/100)n, where i% is the inflation rate. Approximately, the real interest rate ≈ nominal rate − inflation rate; the exact real rate is (1 + r)/(1 + i) − 1.

WORKED EXAMPLE 21 Real value (exam level)

$12 000 is invested for 8 years at 5% compounded annually. Inflation averages 2.5% per year. Find the real value in today's money.

Solution

FV = 12 000(1.05)8 = 17 729.47A1

Real value = 17 729.47/(1.025)8 = $14 551.40M1 A1

Practice exercise 12

1. A price of $80 rises with 3% inflation. Find the price in 10 years.

2. Find the exact real rate when the nominal rate is 6% and inflation 4%.

14 Loans and repayments

Paying back with interest.

A loan of PV is repaid by equal payments PMT at the end of each period. The TVM solver finds any one of N, I%, PV, PMT and FV given the others. For a loan, PV is positive (money received), PMT negative and FV = 0. The total interest is N × |PMT| − PV.

WORKED EXAMPLE 22 Car loan (standard)

A $26 000 car loan at 7.2% compounded monthly is repaid monthly over 5 years. Find the payment and total interest.

Solution

N = 60, I% = 7.2, PV = 26 000, FV = 0, P/Y = C/Y = 12M1

PMT = −$517.29A1

Interest = 60 × 517.29 − 26 000 = $5037.40A1

Practice exercise 13

1. Find the monthly payment on a $12 000 loan at 9% compounded monthly over 4 years.

2. Find the total interest on that loan.

15 Mortgages and the effect of the term

Longer terms, more interest.

Increasing the term of a loan reduces each payment but increases the total interest. Comparing terms is a common examination task.

(Values in the table are for a $200 000 mortgage at 6.5% compounded monthly.)

Term (years)	Monthly payment	Total interest
15	$1742.21	$113 599
20	$1491.15	$157 875
25	$1350.41	$205 124
30	$1264.14	$255 089

WORKED EXAMPLE 23 Choosing a term (exam level)

For the $200 000 mortgage, find the extra interest paid by choosing 30 years rather than 20 years.

Solution

255 089 − 157 875M1

= $97 214A1

Practice exercise 14

1. For the same mortgage over 20 years, find the balance after 5 years.

16 Savings plans and annuities

Regular deposits.

Regular deposits PMT at the end of each period grow to a future value FV. In the TVM solver, PV = 0, PMT is negative (money paid in) and FV is positive. An annuity paid to you from a lump sum has PV negative (you invest it) and PMT positive (you receive payments).

WORKED EXAMPLE 24 Retirement saving (standard)

$400 is saved at the end of each month for 25 years at 5% compounded monthly. Find the final value.

Solution

N = 300, I% = 5, PV = 0, PMT = −400M1

FV = $238 204A1

WORKED EXAMPLE 25 Annuity payout (exam level)

A lump sum of $250 000 is invested at 4% compounded monthly to provide monthly payments for 20 years. Find the payment.

Solution

N = 240, PV = −250 000, FV = 0M1

PMT = $1514.95A1

Practice exercise 15

1. How much must be saved each month for 10 years at 3% compounded monthly to reach $30 000?

2. How long will $100 000 at 5% compounded monthly last with withdrawals of $1000 per month?

17 Finding the number of payments or the rate

Solving for N or I%.

The TVM solver can also find N or I%. A non-integer N means the final payment is smaller than the others. Finding I% shows the true cost of a loan with fees or a payment plan.

WORKED EXAMPLE 26 Finding the rate (exam level)

A shop offers a $2400 television for 24 monthly payments of $112. Find the annual interest rate (compounded monthly).

Solution

N = 24, PV = 2400, PMT = −112, FV = 0M1

I% = 11.1%A1

Practice exercise 16

1. A $15 000 loan at 6% compounded monthly is repaid at $350 per month. Find the number of payments.

2. Find the rate for a $5000 loan repaid by 36 monthly payments of $165.

18 Investment decisions

Comparing options.

Real decisions compare options with different rates, terms and fees. Convert each option to a common basis (effective rate, total cost or future value at the same date) before comparing, and consider risk and inflation.

WORKED EXAMPLE 27 Lump sum or payments (exam level)

A prize is offered as $50 000 now or $600 per month for 8 years. With an interest rate of 5% compounded monthly, which is worth more today?

Solution

PV of payments: N = 96, PMT = 600, I% = 5: PV = $47 394M1 A1

$50 000 now is worth moreR1

Practice exercise 17

1. Option A: $20 000 now. Option B: $24 000 in 4 years. Which is better if money earns 4% compounded annually?

19 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 28 Mixed problem 1 (exam level)

Sara borrows $18 000 at 8% compounded monthly and repays $450 per month. (a) How many payments are needed? (b) What is the balance after 2 years?

Solution

(a) N = 46.7: 47 payments (the last one smaller)M1 A1

(b) N = 24: balance $9442 (approximately)M1 A1

WORKED EXAMPLE 29 Mixed problem 2 (exam level)

An investment of $8000 doubles in 9 years with monthly compounding. Find the nominal annual rate.

Solution

(1 + r/1200)108 = 2M1

r = 7.73%A1

Practice exercise 18

1. Find the effective rate for 12% compounded monthly.

2. A $300 000 mortgage at 5% compounded monthly over 25 years: find the payment.

20 Additional exam-style practice

Questions in the style of Paper 2.

WORKED EXAMPLE 30 Paper 2 question (exam level)

Tom invests $5000 at 3.5% compounded quarterly. Jan invests $5000 at 3.6% compounded annually. (a) Find each value after 10 years. (b) Find the first full year in which Jan's value exceeds Tom's.

Solution

(a) Tom $7084.54; Jan $7121.44A1 A1

(b) Compare yearly: Jan's effective rate 3.6% > Tom's 3.546%, so Jan is ahead from year 1R1 A1

Practice exercise 19

1. A car costs $32 000. Option 1: pay cash with a 5% discount. Option 2: a loan at 3.9% compounded monthly over 4 years. Find the total cost of option 2 and compare.

21 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 2 style) (about 15 minutes)

Option A: $12 000 invested at 5% per annum compounded annually. Option B: $12 000 invested at 4.8% per annum compounded monthly.

(a) Find the value of option A after 10 years. [2]

(b) Find the value of option B after 10 years. [2]

(d) Inflation is 2% per year. Estimate the real value of option A after 10 years. [2]

Markscheme

Part	Answer	Marks
(a)	12 000 × 1.0510	M1
	$19 546.74	A1
(b)	12 000 × (1 + 4.8/1200)120	M1
	$19 374.33	A1
(c)	12 000 × 1.05n > 25 000	M1
	n > 15.04	A1
	16 years	A1
(d)	12 000 × 1.0310	M1
	$16 127.00	A1

Examiner's comments. In (c), the integer answer must be given; 15.04 years would lose the final mark.

Question 2 (Paper 2 style) (about 12 minutes)

Mei borrows $18 000 to buy a car. The loan is repaid with equal monthly payments over 4 years at 6.6% per annum compounded monthly.

(a) Find the monthly payment. [3]

(b) Find the total interest paid. [2]

Markscheme

Part	Answer	Marks
(a)	N = 48, I% = 6.6, PV = 18 000, FV = 0, P/Y = C/Y = 12	M1
	PMT = −427.70	A1
	$427.70	A1
(b)	48 × 427.70 − 18 000	M1
	$2529.60	A1
(c)	N = 24 with same PV and PMT; solve FV	M1
	$9591.51	A1

Examiner's comments. Part (c) requires recognizing that the outstanding balance is the future value after 24 payments.

22 Common misconceptions and links

Students often think…	Actually…
more frequent compounding doubles interest	the gain is small
depreciation uses 1 + r/100	use 1 − r/100
N is the number of years	N is the number of periods
PV and PMT have the same sign for a loan	borrowed is positive, repayments negative

How 1.4 connects to the rest of the course

Topic	Connection
1.1 Sequences and series	geometric sequences
1.5 Exponents and logarithms	solving for n
2.8 Exponential functions	continuous growth
5.18 Differential equations (HL)	continuous models

23 Check your understanding

Section A: multiple choice

1. $1000 at 10% compound interest for 2 years becomes

A. $1210 B. $1200 C. $1100 D. $1221

2. The multiplier for a 7% depreciation is

A. 1.07 B. 0.07 C. 0.93 D. 0.7

3. For quarterly compounding over 3 years, N =

A. 3 B. 12 C. 4 D. 36

4. A deposit in the TVM solver is entered as

A. a positive PV B. PMT C. FV D. a negative PV

5. Real rate with 5% interest and 3% inflation ≈

A. 8% B. 2% C. 15% D. 1.67%

6. Compound interest balances form

A. a geometric sequence B. an arithmetic sequence C. a constant sequence D. a quadratic sequence

Section B: structured

7. Compare simple and compound interest on $2000 at 6% for 15 years. [3]

8. Find the annual rate needed to double money in 12 years. [2]

9. Explain why the TVM sign convention matters. [2]

Answers

1. A 2. C 3. B 4. D 5. B 6. A

7. Simple: $3800; compound: $4793.12; compound is larger by $993.12.

8. 21/12 = 1.0595: 5.95%.

9. Money in and out must have opposite signs; otherwise the solver cannot find a solution.

24 Summary and topic map

Idea	Remember
Simple vs compound	arithmetic vs geometric growth
Compound interest	FV = PV(1 + r/100k)kn
Effective rate	(1 + r/100k)k − 1
Depreciation	V = V0(1 − r/100)n
Inflation	real rate ≈ nominal − inflation
TVM	N, I%, PV, PMT, FV, P/Y, C/Y; sign convention
Loans	PV positive, PMT negative, FV 0

Before the exam, make sure you…

•use kn periods and r/k per period
•give money to 2 d.p.
•round years up when a target must be reached
•list TVM inputs in your answer

Financial Applications on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 2. Topic map for Financial Applications.

25 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. $504

2. $2950.21

3. arithmetic: the same amount is added each year

Practice exercise 2

1. $798.60

2. $20 771.76

3. $2791.97

Practice exercise 3

1. $3379.48

2. 8.24%

3. $2321.08

4. 5.85% monthly gives 6.01% effective: slightly better

Practice exercise 4

1. 11.9

2. 9.01, so 10 years

3. 5.36%

4. 33 quarters

Practice exercise 5

1. $411.60

2. 9 years (n > 8.31)

3. 14.8%

Practice exercise 6

1. 2.5%

2. $5523.11

3. $6.81

Practice exercise 7

1. $362.69

2. 48 × 362.68 − 15 000 = $2408.64

3. $61 655.05

4. $365.48

Practice exercise 8

1. $4239.70

2. 16.2, so 17 years

3. 5.20%

Practice exercise 9

1. $3590.68

2. $744.24

Practice exercise 10

1. 7.19%

2. 4.89%

Practice exercise 11

1. $675

2. $6144 vs $3000

Practice exercise 12

1. $107.51

2. 1.92%

Practice exercise 13

1. $298.62

2. $2333.76

Practice exercise 14

1. N = 60: balance $171 178 (approximately)

Practice exercise 15

1. $214.69

2. N = 129.6, so about 130 months (10.8 years)

Practice exercise 16

1. N = 48.4 (49 payments, the last smaller)

2. I% = 11.6%

Practice exercise 17

1. PV of B = 24 000/1.044 = $20 515: B is better

Practice exercise 18

1. 12.68%

2. $1753.77

Practice exercise 19

1. PMT = $721.11; total $34 613 vs $30 400: cash is cheaper

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