DP Maths AI SL 1.3 Geometric Sequences and Series

DP Maths AI SL 1.3 Geometric Sequences and Series - Full notes

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DP Maths AI SL | 1.3 Geometric Sequences and Series

DP MATHEMATICS AI · TOPIC 1: NUMBER AND ALGEBRA

1.3

Geometric Sequences and Series

Full notes for Applications and Interpretation SL

Contents

1	Geometric patterns
2	The common ratio and the nth term
3	Finding unknown terms and ratios
4	Finding n with logarithms and technology
5	Geometric series
6	Sums with sigma notation and technology
7	Percentage growth and decay
8	Compound interest
9	Compounding periods and effective rates
10	Depreciation
11	Inflation and real value
12	The financial solver: loans and annuities
13	Modelling populations and other quantities
14	Comparing arithmetic and geometric models
15	Deciding whether data are geometric
16	Spreadsheets and technology
17	Mixed review: worked problems
18	Investigation: the bouncing ball
19	Exam-style questions with markscheme
20	Common misconceptions and links
21	Check your understanding
22	Summary and topic map
23	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 practice exercise.

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
identify geometric sequences and find the common ratio
use un = u1rn−1 to find terms, ratios and positions
find n using logarithms, tables or solvers
use Sn = u1(rn − 1)/(r − 1)
model percentage growth and decay
solve compound interest, depreciation and inflation problems
use the financial (TVM) solver for loans and annuities
decide whether data are approximately geometric

SL booklet: this booklet contains the content for Applications and Interpretation SL. Sections marked for HL in the HL edition are not included.

Guiding questions

How can quantities that grow or shrink by a constant percentage be modelled? How do banks calculate interest, loans and savings?

How this topic is examined

Geometric sequences and financial mathematics are tested on both papers, usually in context: populations, compound interest, depreciation, loan repayments and savings plans. Paper 2 expects efficient use of the TVM solver and tables.

Convert percentages to multipliers (a 4% increase means r = 1.04; a 15% decrease means r = 0.85), show the formula or the TVM inputs you used, and give money to two decimal places.

Mathematics: Applications and Interpretation places strong emphasis on technology, modelling and interpretation. Most papers allow a GDC, so you are expected to use it efficiently, record the method you used, and interpret every answer in context.

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, including what you entered into a GDC
Show that	obtain the given result; every step must be shown and the result must not be assumed
Estimate	obtain an approximate value
Sketch	a diagram showing the general shape and key features, labelled
Interpret / Comment	explain the meaning of a result in context, or give a judgement
Justify / Explain	give valid reasons or evidence
Hence	use the previous result

How to use these notes

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or explanation of a key result
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
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), graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet

1 Geometric patterns

Multiply by the same number each time.

Think about it

A rumour is told to 3 people, each of whom tells 3 new people the next hour, and so on. How many people hear it in the 5th hour? How does this compare with adding 3 people each hour?

A sequence is geometric if each term is obtained by multiplying the previous term by the same non-zero constant, the common ratio r. Equivalently, the ratio of consecutive terms is constant: r = un+1/un.

The sequence 3, 9, 27, 81, ... has r = 3; 400, 200, 100, 50, ... has r = 0.5; 5, −10, 20, −40, ... has r = −2 and alternates in sign. When r > 1 the terms grow; when 0 < r < 1 they shrink towards zero; when r < 0 they alternate.

Geometric sequences model multiplicative change: quantities that change by a fixed percentage, such as compound interest, population growth, radioactive decay and depreciation.

Sequence	Ratio	Behaviour
2, 6, 18, 54	3	growth
1000, 900, 810	0.9	decay
7, −14, 28	−2	alternating, growing
64, −32, 16	−0.5	alternating, shrinking

WORKED EXAMPLE 1 Identifying a geometric sequence (basic)

Determine whether each is geometric: (a) 5, 15, 45, 135; (b) 2, 4, 8, 14.

Solution

(a) ratios 3, 3, 3: geometric, r = 3A1

(b) ratios 2, 2, 1.75: not geometricA1

Practice exercise 1

1. Write the next two terms of 0.5, 2, 8, ... .

2. Find r for 81, −27, 9, ... .

3. Is 10, 12, 14.4, 17.28 geometric?

4. Write the first four terms of a geometric sequence with u1 = 6 and r = −0.5.

2 The common ratio and the nth term

un = u1rn−1.

From u1, each step multiplies by r, so the nth term has been multiplied (n − 1) times. The recursive form is un+1 = r un. Plotted against n, the terms lie on an exponential curve, not a straight line.

Figure 1. un = 500(0.8)n−1: geometric decay.

WORKED EXAMPLE 2 Finding a term (basic)

Find u12 for 4, 6, 9, ... .

Solution

r = 1.5A1

u12 = 4(1.5)11 = 346 (3 s.f.)A1

WORKED EXAMPLE 3 Alternating sequence (standard)

Find u9 for 3, −6, 12, ... .

Solution

r = −2A1

u9 = 3(−2)8 = 768A1

Common mistake

Using rn instead of rn−1, or computing (u1r)n−1 instead of u1 × rn−1.

Practice exercise 2

1. Find u8 for 2, 10, 50, ... .

2. Find u15 for 1000, 950, 902.5, ... .

3. Write an explicit formula for u1 = 12, un+1 = 0.75un.

3 Finding unknown terms and ratios

Divide one term by another.

If two terms are given, dividing them eliminates u1: um/uk = rm−k. Take the appropriate root to find r (remember that an even root gives ±). For three consecutive terms a, b, c, the condition b2 = ac (b is the geometric mean) gives an equation for an unknown.

WORKED EXAMPLE 4 Two terms (standard)

In a geometric sequence u3 = 20 and u6 = 160. Find r and u1.

Solution

r3 = 160/20 = 8M1

r = 2A1

u1 = 20/4 = 5A1

WORKED EXAMPLE 5 Even power (exam level)

u2 = 12 and u4 = 108. Find the possible values of r and u1.

Solution

r2 = 9M1

r = 3, u1 = 4 or r = −3, u1 = −4A1 A1

WORKED EXAMPLE 6 Consecutive terms (exam level)

k, k + 6 and 4k are consecutive terms of a geometric sequence with k > 0. Find k.

Solution

(k + 6)2 = 4k2M1

k + 6 = 2k (positive root)A1

k = 6 (terms 6, 12, 24)A1

Practice exercise 3

1. u4 = 54 and u7 = 1458. Find r and u1.

2. u3 = 2.5 and u5 = 0.625. Find the possible values of r.

3. Find x > 0 if x − 2, x + 1, 3x + 3 are consecutive terms of a geometric sequence.

4 Finding n with logarithms and technology

The unknown in the exponent.

When n is unknown, the equation u1rn−1 = value has n in the exponent. Solve it with logarithms (n − 1 = log(value/u1)/log r), with a GDC table, or with the equation solver. For inequalities, remember that dividing by log r reverses the inequality when 0 < r < 1 (because log r < 0); a table avoids this trap.

WORKED EXAMPLE 7 First term above a value (standard)

In 5, 7.5, 11.25, ..., which is the first term greater than 1000?

Solution

5(1.5)n−1 > 1000, (1.5)n−1 > 200M1

n − 1 > ln 200/ln 1.5 = 13.07A1

n = 15A1

WORKED EXAMPLE 8 Decay below a value (standard)

A drug's concentration falls by 18% per hour from 60 mg/L. After how many whole hours is it first below 5 mg/L?

Solution

60(0.82)t < 5M1

t > ln(1/12)/ln 0.82 = 12.52A1

13 hoursA1

Practice exercise 4

1. Which term of 3, 6, 12, ... is 6144?

2. How many terms of 2000, 1600, 1280, ... are greater than 100?

3. A population of 850 grows by 6% per year. In which year does it first exceed 2000?

5 Geometric series

Summing the terms.

Why it works

Write Sn = u1 + u1r + ... + u1rn−1 and rSn = u1r + ... + u1rn. Subtracting, all terms but two cancel: rSn − Sn = u1rn − u1, so Sn(r − 1) = u1(rn − 1).

WORKED EXAMPLE 9 Sum of terms (basic)

Find the sum of the first 10 terms of 3, 6, 12, ... .

Solution

S10 = 3(210 − 1)/(2 − 1)M1

= 3069A1

WORKED EXAMPLE 10 Decaying series (standard)

A well produces 5000 barrels in month 1, and production falls by 4% each month. Find the total production in the first two years.

Solution

S24 = 5000(1 − 0.9624)/0.04M1

= 78 100 barrels (3 s.f.)A1

WORKED EXAMPLE 11 Finding the ratio (exam level)

A geometric series has u1 = 4 and S3 = 28. Find the possible values of r.

Solution

4(1 + r + r2) = 28M1

r2 + r − 6 = 0A1

r = 2 or r = −3A1

Practice exercise 5

1. Find S8 for 1, 3, 9, ... .

2. Find the sum of the first 12 terms of 256, 192, 144, ... .

3. Find the least n for which Sn of 2 + 4 + 8 + ... exceeds 10 000.

6 Sums with sigma notation and technology

Σ for geometric series.

A geometric series can be written as Σ from k = 1 to n of u1rk−1. Identify the first term by substituting the lower limit, the ratio from the base, and count the terms carefully. On a GDC, use the sum template or sum(seq(...)).

WORKED EXAMPLE 12 Sigma notation (standard)

Evaluate Σ from k = 1 to 8 of 5(1.2)k.

Solution

First term 6, r = 1.2, 8 termsM1

S8 = 6(1.28 − 1)/0.2 = 99.0A1

Practice exercise 6

1. Evaluate Σ from k = 0 to 6 of 2k.

2. Write 3 + 12 + 48 + ... + 3072 in sigma notation.

7 Percentage growth and decay

Multipliers.

An increase of p% multiplies by (1 + p/100); a decrease of p% multiplies by (1 − p/100). After n periods, the value is V0(1 ± p/100)n. Note the indexing: if V0 is the initial value, Vn is the value after n periods (the (n + 1)th term).

Change	Multiplier
+5%	1.05
+0.8%	1.008
−12%	0.88
−1.5%	0.985
doubles	2

WORKED EXAMPLE 13 Successive percentage changes (standard)

A price rises by 10% and then falls by 10%. Find the overall percentage change.

Solution

1.1 × 0.9 = 0.99M1

1% decreaseA1

Common mistake

Adding percentages: a 10% rise followed by a 10% fall is not 0% change.

Practice exercise 7

1. A town of 24 000 grows by 2.5% per year. Find its population after 8 years.

2. A car loses 15% of its value each year. What percentage of its value remains after 4 years?

3. Find the single multiplier equivalent to a 6% rise followed by a 4% rise.

8 Compound interest

Interest on interest.

PV is the present value (principal), r the nominal annual rate (%), k the number of compounding periods per year and n the number of years. With annual compounding (k = 1), the balances form a geometric sequence with ratio 1 + r/100.

WORKED EXAMPLE 14 Annual compounding (basic)

$3500 is invested at 4.2% per year compounded annually. Find the value after 6 years.

Solution

3500(1.042)6M1

= $4479.96A1

WORKED EXAMPLE 15 Finding the rate (standard)

An investment of $10 000 grows to $13 000 in 5 years with annual compounding. Find the rate.

Solution

(1 + r/100)5 = 1.3M1

r = 100(1.30.2 − 1) = 5.39%A1

Practice exercise 8

1. Find the value of $800 after 10 years at 6% compounded annually.

2. How long does it take money to double at 3% compounded annually?

3. Find the interest earned on $5000 at 2.8% for 7 years compounded annually.

9 Compounding periods and effective rates

Monthly, quarterly, daily.

With k compounding periods per year, the rate per period is r/k % and there are kn periods. More frequent compounding gives slightly more interest. To compare offers, compute the effective annual rate, (1 + r/(100k))k − 1.

WORKED EXAMPLE 16 Monthly compounding (standard)

$4000 at 3.6% per year compounded monthly for 5 years.

Solution

4000(1 + 3.6/1200)60M1

= $4787.58A1

WORKED EXAMPLE 17 Comparing offers (exam level)

Which is better: 4.5% compounded annually or 4.4% compounded monthly?

Solution

Annual: 4.5%; monthly: (1 + 0.044/12)12 − 1 = 4.49%M1 A1

4.5% annually is very slightly betterR1

Practice exercise 9

1. Find the value of $2000 after 3 years at 6% compounded quarterly.

2. Find the effective annual rate of 5% compounded daily.

10 Depreciation

Losing a fixed percentage.

Depreciation at r% per year multiplies the value by (1 − r/100) each year: V = V0(1 − r/100)n. Unlike straight-line depreciation (arithmetic), the value never reaches zero.

WORKED EXAMPLE 18 Depreciation (standard)

A van bought for $36 000 depreciates by 14% per year. Find its value after 5 years and when it first falls below $10 000.

Solution

36 000(0.86)5 = $16 935.37A1

0.86n < 10/36: n > 8.49M1

After 9 yearsA1

Practice exercise 10

1. A phone costing $950 depreciates by 30% per year. Find its value after 3 years.

2. A $25 000 car is worth $12 000 after 4 years. Find the annual depreciation rate.

11 Inflation and real value

Purchasing power.

Inflation is the general rise in prices. To find what money is worth in today's terms, adjust for inflation. A simple approximation used in AI: real rate ≈ nominal rate − inflation rate. To find a future price, grow today's price at the inflation rate.

WORKED EXAMPLE 19 Real value (standard)

$20 000 is invested at 5% compounded annually for 10 years, while inflation averages 2.2%. Estimate the real value.

Solution

Real rate ≈ 2.8%M1

20 000(1.028)10 = $26 360.96A1

Practice exercise 11

1. A loaf costs $3.50 today. With 3% inflation, estimate its cost in 12 years.

2. Savings earn 1.5% while inflation is 4%. Describe what happens to their real value.

12 The financial solver: loans and annuities

Regular payments.

The TVM solver handles regular payments. Inputs: N (number of payments), I% (annual rate), PV, PMT, FV, P/Y (payments per year) and C/Y (compounding periods per year). Sign convention: money you pay out is negative, money you receive is positive. For a loan, PV > 0 (you receive it), PMT < 0, FV = 0. For savings, PMT < 0 and FV > 0.

Situation	PV	PMT	FV
Single investment	−	0	+
Loan repaid in full	+	−	0
Regular savings	0	−	+

WORKED EXAMPLE 20 Loan repayment (standard)

A $15 000 loan at 6.5% per year compounded monthly is repaid monthly over 4 years. Find the payment and the total interest.

Solution

N = 48, I% = 6.5, PV = 15 000, FV = 0, P/Y = C/Y = 12M1

PMT = −$355.72A1

Total interest = 48 × 355.72 − 15 000 = $2074.56A1

WORKED EXAMPLE 21 Savings plan (standard)

$250 is deposited at the end of each month at 4% compounded monthly. Find the value after 15 years.

Solution

N = 180, I% = 4, PV = 0, PMT = −250M1

FV = $61 522.62A1

WORKED EXAMPLE 22 Number of payments (exam level)

A $200 000 mortgage at 5.4% compounded monthly is repaid at $1500 per month. How long does it take?

Solution

I% = 5.4, PV = 200 000, PMT = −1500, FV = 0M1

N = 204.1A1

205 months (about 17.1 years)A1

GDC tip

If the solver gives an error, check that PV and FV (or PV and PMT) have opposite signs, and that P/Y and C/Y are set before solving.

Practice exercise 12

1. Find the monthly repayment on a $9000 loan over 3 years at 8% compounded monthly.

2. How much must be saved each month for 10 years at 3.5% compounded monthly to reach $50 000?

3. A $30 000 loan at 7% compounded annually is repaid with annual payments of $5000. How many years?

13 Modelling populations and other quantities

Exponential growth in discrete steps.

A population increasing by a constant percentage each period follows a geometric model Pn = P0(1 + p/100)n. Such models are reasonable over short periods but ignore limits on resources; real populations eventually level off (logistic models, HL).

WORKED EXAMPLE 23 Bacteria (standard)

A culture of 500 bacteria doubles every 3 hours. Find the number after 24 hours and when it first exceeds one million.

Solution

8 doublings: 500 × 28 = 128 000A1

500 × 2t/3 > 106: t > 32.9M1

After 33 hoursA1

Practice exercise 13

1. A species of fish declines by 7% per year from 12 000. Find the population after 10 years.

2. Explain one limitation of a geometric population model.

14 Comparing arithmetic and geometric models

Linear versus exponential change.

Arithmetic models add a constant; geometric models multiply by a constant. Over a short time they may look similar, but geometric growth (r > 1) eventually overtakes any arithmetic growth.

Figure 2. Geometric growth overtakes arithmetic growth.

WORKED EXAMPLE 24 Overtaking (exam level)

Plan A pays $1000 in year 1, rising by $150 per year; plan B pays $1000 rising by 10% per year. In which year does B first pay more than A?

Solution

1000(1.1)n−1 > 1000 + 150(n − 1)M1

Table: n = 7: 1771.56 vs 1900; n = 8: 1948.72 vs 2050; n = 9: 2143.59 vs 2200; n = 10: 2357.95 vs 2350A1

Year 10A1

Practice exercise 14

1. Which is larger after 30 years: $5000 at 6% simple interest or $5000 at 5% compound interest?

15 Deciding whether data are geometric

Check the ratios.

If the ratios of consecutive values are roughly constant, a geometric (exponential) model is appropriate. The average ratio estimates r. In HL, taking logarithms of geometric data gives a straight line (see the logarithms booklet), which allows linear regression.

Year	2020	2021	2022	2023	2024
Users (thousands)	12.0	15.1	18.8	23.6	29.4
Ratio		1.258	1.245	1.255	1.246

WORKED EXAMPLE 25 Fitting a geometric model (exam level)

Estimate r from the table, write a model and predict users in 2027.

Solution

Mean ratio ≈ 1.25M1 A1

Un = 12(1.25)n, n years after 2020A1

2027: 12(1.25)7 = 57.2 thousand; extrapolation, so uncertainA1 R1

Practice exercise 15

1. Values 80, 60, 45.2, 33.8 are measured hourly. Is a geometric model reasonable? Estimate r.

16 Spreadsheets and technology

Tables, lists and recursion.

A spreadsheet column with =B1*r generates a geometric sequence; a running total column gives Sn. The GDC sequence mode u(n) = r u(n − 1) works the same way. Tables are the most reliable way to answer 'first n such that' questions and avoid inequality sign errors with logarithms.

n	un (=B1*1.08)	Sn
1	500	500
2	540	1040
3	583.20	1623.20
4	629.86	2253.06
5	680.24	2933.30

Practice exercise 16

1. Use a table to find the first n for which 500(1.08)n−1 exceeds 2000.

2. Use a running total to find S15 for u1 = 500, r = 1.08.

17 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 26 Mixed problem 1 (standard)

The 2nd term of a geometric sequence is 18 and the 5th term is 486. Find the sum of the first 6 terms.

Solution

r3 = 27, r = 3, u1 = 6M1 A1

S6 = 6(36 − 1)/2 = 2184A1

WORKED EXAMPLE 27 Mixed problem 2 (exam level)

Mia invests $5000 at 3.8% compounded quarterly. Leo invests $5500 at 3.2% compounded annually. After how many whole years does Mia's investment first exceed Leo's?

Solution

5000(1.0095)4n > 5500(1.032)nM1

Table: n = 15: 8818 vs 8822; n = 16: 9157 vs 9104A1

16 yearsA1

WORKED EXAMPLE 28 Mixed problem 3 (exam level)

A ball dropped from 2 m rebounds to 65% of its previous height. Find the total vertical distance travelled when it hits the ground for the 8th time.

Solution

Down: 2 + 2(0.65 + ... + 0.657)M1

Up: 2(0.65 + ... + 0.657) included above; total = 2 + 2 × 2 × 0.65(1 − 0.657)/0.35M1

= 9.06 mA1

Practice exercise 17

1. A geometric series has S2 = 15 and S4 = 255. Find r (r > 0) and u1.

2. An investment doubles in 9 years with annual compounding. Find the interest rate.

18 Investigation: the bouncing ball

A Paper 3 style exploration.

A ball is dropped from height h and rebounds to a fraction e of its previous height each time (e is called the coefficient of restitution squared, in physics terms).

1. Show that the heights reached after each bounce form a geometric sequence and state its ratio. 2. Find the total distance travelled up to the nth impact. 3. Using a GDC, find how many bounces occur before the rebound height is below 1 cm, for h = 2 m and e = 0.7. 4. (HL) Find the total distance travelled before the ball comes to rest. 5. Comment on the realism of the model.

Answers. 1. Heights he, he2, ...: ratio e. 2. h + 2he(1 − en−1)/(1 − e). 3. 2(0.7)n < 0.01: n > 14.9, so 15 bounces. 4. h(1 + e)/(1 − e) = 2(1.7)/0.3 = 11.3 m. 5. Real balls lose energy differently at low speed and stop after a finite time.

19 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 10 minutes)

The first three terms of a geometric sequence are 48, 36 and 27.

(a) Find the common ratio. [1]

(b) Find u8. [2]

Markscheme

Part	Answer	Marks
(a)	0.75	A1
(b)	48(0.75)7	M1
	6.41	A1
(c)	48(1 − 0.758)/0.25	M1
	172	A1

Examiner's comments. Give answers to 3 s.f. unless exact.

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

Kai borrows $24 000 to buy a car at 5.9% per year compounded monthly, repaid by equal monthly payments over 5 years.

(a) Find the monthly payment. [3]

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

(d) Find the balance of the loan after 3 years. [2]

Markscheme

Part	Answer	Marks
(a)	N = 60, I% = 5.9, PV = 24 000, FV = 0, P/Y = C/Y = 12	M1
	PMT = −462.87	A1
	$462.87	A1
(b)	60 × 462.87 − 24 000	M1
	$3772.20	A1
(c)	24 000(0.83)5	M1
	$9453.70	A1
(d)	N = 36, solve FV	M1
	$10 454	A1

Examiner's comments. List the TVM inputs; answers without inputs earn limited credit.

Question 3 (Paper 2 style) (about 10 minutes)

A town's population was 18 500 at the start of 2020 and grows by 2.4% per year.

(a) Find the population at the start of 2030. [2]

(b) Find the year in which the population first exceeds 30 000. [3]

Markscheme

Part	Answer	Marks
(a)	18 500(1.024)10	M1
	23 452	A1
(b)	18 500(1.024)n > 30 000	M1
	n > 20.4	A1
	during 2040 (start of 2041)	A1
(c)	long-term extrapolation; growth rates change	R1

Examiner's comments. State clearly whether the answer refers to the start or during the year.

20 Common misconceptions and links

Students often think…	Actually…
a 5% increase means r = 0.05	r = 1.05
un = u1rn	un = u1rn−1
PV and FV have the same sign	money in and out have opposite signs
dividing by log 0.9 keeps the inequality	log 0.9 < 0 reverses it

How 1.3 connects to the rest of the course

Topic	Connection
1.2 Arithmetic sequences	constant difference
1.5 Exponential functions	continuous growth
1.9 Logarithms	solving for n
5.14 Differential equations	continuous models

21 Check your understanding

Section A: multiple choice

1. The ratio of 50, 40, 32, ... is

A. 0.8 B. −10 C. 1.25 D. 0.2

2. u5 of 2, 6, 18, ... is

A. 486 B. 54 C. 162 D. 30

3. A 12% decrease uses multiplier

A. 1.12 B. 0.88 C. 0.12 D. 0.8

4. For a loan in the TVM solver, PV is

A. negative B. zero C. equal to FV D. positive

5. Compound interest balances form

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

Section B: structured

6. $6000 is invested at 4% compounded monthly. Find the number of whole months until it exceeds $8000. [3]

7. Explain why depreciation by a fixed percentage never reaches zero. [1]

Answers

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

6. 6000(1 + 0.04/12)m > 8000: m > 86.4, so 87 months.

7. multiplying a positive value by 0.85 always gives a positive value.

22 Summary and topic map

Idea	Remember
Geometric	constant ratio r
nth term	u1rn−1
Sum	u1(rn − 1)/(r − 1)
Growth/decay	multiplier 1 ± p/100
Compound interest	PV(1 + r/100k)kn
Depreciation	V0(1 − r/100)n
TVM	N, I%, PV, PMT, FV, P/Y, C/Y; sign convention

Before the exam, make sure you…

•convert percentages to multipliers
•use tables for 'first n' questions
•list TVM inputs
•interpret in context and note limitations

Geometric 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 3. Topic map for Geometric Sequences and Series.

23 Answers to practice exercises

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

Practice exercise 1

1. 32, 128

2. −1/3

3. yes, r = 1.2

4. 6, −3, 1.5, −0.75

Practice exercise 2

1. 156 250

2. 1000(0.95)14 = 488

3. un = 12(0.75)n−1

Practice exercise 3

1. r = 3, u1 = 2

2. ±0.5

3. (x + 1)2 = (x − 2)(3x + 3): x2 + 2x + 1 = 3x2 − 3x − 6, 2x2 − 5x − 7 = 0, x = 3.5

Practice exercise 4

1. 12th

2. un > 100 for n ≤ 14: 14 terms

3. 850(1.06)n > 2000: n > 14.7, during year 15

Practice exercise 5

1. 3280

2. 256(1 − 0.7512)/0.25 = 991

3. 2(2n − 1) > 10 000: n = 13

Practice exercise 6

1. 127

2. Σ from k = 1 to 6 of 3(4)k−1

Practice exercise 7

1. 29 242

2. 52.2%

3. 1.1024

Practice exercise 8

1. $1432.68

2. 23.4, so 24 years

3. $1066.27

Practice exercise 9

1. $2391.24

2. 5.13%

Practice exercise 10

1. $325.85

2. 16.8%

Practice exercise 11

1. $4.99

2. real value falls by about 2.5% per year

Practice exercise 12

1. $282.03

2. $348.60

3. N = 8.05, so 9 payments (the last one smaller)

Practice exercise 13

1. 5808

2. it grows without limit; real populations are limited by resources

Practice exercise 14

1. simple: $14 000; compound: $21 609.71: compound larger

Practice exercise 15

1. ratios 0.75, 0.753, 0.748: yes, r ≈ 0.75

Practice exercise 16

1. u19 = 1998.0, u20 = 2157.9: n = 20

2. 13 576.06

Practice exercise 17

1. S4/S2 = 1 + r2 = 17: r = 4, u1 = 3

2. 8.01%

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