ARITHMETIC

AND

GEOMETRIC

SEQUENCES

MATHEMATICS3 ESO

Sequences (Sucesiones)

A number sequence is a set of numbers, usually separated by commas, arranged in an

order. The first term is referred to as t 1 , the second term as t 2 , the third term as t 3 and so on. The nth term is referred to as t n .

A sequence may stop at some number, or continue indefinitely. The following are examples of sequences:

a) 5, 7, 9, 11 t 1=5 ; t 2=7 ; t 3=9 ; t 4=11b) 80, 40, 20, 10, The three dots, called an ellipsis, indicate that the sequence continues indefinitely.

Exercise

1.- Describe the pattern and write the next three terms.

a) 4, 7, 10, 13, pattern: _________________________________________________

b) 3, 6, 12, 24, pattern: _________________________________________________

c) 1, 2, 4, 7, 11, 16, pattern: _________________________________________________

d) 4, 12, 6, 18, 9, pattern: _________________________________________________

e) 2, 3, 9, 10, 30, pattern: _________________________________________________

f) 3, 4, 7, 11, 18, 29, pattern: _________________________________________________

g) 9, 27 45, 63, pattern: _________________________________________________

h) 1, 4, 9, 16, 25, pattern: _________________________________________________

k) 2, 5, 10, 17, 26, pattern: _________________________________________________

l) 11, 8, 3, 5, -2, 7, pattern: _________________________________________________

2.- Imagine a pattern to create your own sequence. Write down its first five terms.

Finding the formula for the n th term

Sometimes a pattern can lead to a general rule for finding the terms of a sequence. The rule is

called the formula for the nth term, t n . Given the formula for the nth term, t n , the terms of a sequence can be written by substituting term numbers for n.

Practise

1.- Given the formula for the nth term, state the first five terms of the sequence.

a) an=3n a1=31=3 ; a2=32=6 ; a3= ; a4= ; a5= ; ; a18=

b) bn=2 n4

c) cn=52 n

d) d n=2n

e) en=n3n4

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Example

a) Find a formula for the nth term that determines the sequence 4, 7, 10, 13,

b) Use the formula for the nth term to find t 35 and t 50

Solution

a) Find a pattern and use it to write an expression for each term as a function of its term number.

t 1=4 or 311

t 2=7 or 321

t 3=10 or 331

t 4=13 or 341

So, t n=3n1 . The formula for the nth term is tn=3n1

b) t 35=3351=106 ; t 50=3501=151

Practise

2.- Find a formula for the nth term that determines each sequence. Then, list the next three terms.

a) 5, 10, 15, 20, b) 2, 3, 4, 5, ...

c) 6, 5, 4, 3, d) 1, 4, 9, 16,

e) 2, 4, 6, 8, f) -3, -6, -12, -24,

2

g) -1, 0, 1, 2, h) 0.1, 0.2, 0.3, 0.4,

k) 12

,23

,34

,45 , ... l) x, 2x, 3x, 4x,

3.- List the first four terms of the sequence determined by each of the following.

a) t n=n12

b) t n=n1n1

c) t n=1n

d) t n=1

3n11

4.- The I-Shapes are made from asterisks.

a) How many asterisks are there in the fourth diagram?

b) The fifth diagram?

c) Describe the pattern in words.

d) Write an expression that represents the number of asterisks in the nth I-shape in terms of n.

e) Using the expression from question d), how many asterisks are there in the 65th I-shape? The 100th I-shape?

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Arithmetic Sequences (Progresiones aritmticas)

A sequence like 2, 5, 8, 11, , where the difference between consecutive terms is a constant, is called an arithmetic sequence. In an arithmetic sequence, the first term,

t 1 , is denoted by the letter a. Each term after the first is found by adding a constant, called the common difference, d, to the preceding term.

Example 1

Given the formula for the nth term of a sequence, t n=2n1 , write the first 6 terms. Notice that this is an arithmetic sequence. What is the difference for this arithmetic sequence?

Finding the formula for the n th term of a general arithmetic sequence

Note that a general arithmetic sequence where a is the first term and d is the common difference is

t1 = a

t2 = adt 3 = ad d = a2 dt 4 = a2d d = a3d

...

t n = an1d where n is a natural number.Note that d is the difference between any successive pair of terms. For example,

t 2t 1 = ada = dt 3t 2 = a2 dad = a2 dad = d

Example 2. Finding the Formula for the nth Term.

Find the formula for the nth term, t n , for the arithmetic sequence 8, 12, 16,

Solution

For the given sequence, a=8 and d=4t n = an1d

Substitute known values: t n = 8n14

Expand: t n= 84 n4

Simplify: t n= 4n4

Now check the three first terms and find t 4 , t 5 and t 21 using the formula for the nth term.

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Example 3. Finding the Number of Terms

How many terms are there in the following sequence?

3 , 2, 7, ... , 152Solution

For the given sequence, a=3 , d=5 , and t n= 152Substitute the known values in the formula for the general term and solve for n.

t n = an1d

Substitute known values: 152=3n15Expand: 152=35 n5Simplify: 152= 5 n8Solve for n: 1528 = 5 n88

160 = 5 n160

5 =5 n5 32=n

The sequence has 32 terms.

Example 4. Finding t n Given Two Terms

In an arithmetic sequence, t 7= 121 and t 15 = 193 . Find the first three terms of the sequence and tn

Solution

Substitute known values in the formula for the nth term to write a system of equations. Then, solve the system.

t n= an1d

Write an equation for t 7 : 121=a71d 121=a6 d (1)

Write an equation for t 15 : 193=a151d 193=a14 d (2)

Subtract (1) from (2) : 72=8 d d = 9

Substitute 9 for d in (1): 121=a69

Solve for a: 121=a54 a = 67

Since a = 67 and d = 9 , the first three terms of the sequence are 67, 76, and 85.

To find t n , substitute 67 for a and 9 for d in the formula for the nth term.

t n= an1d

t n = 67n19

t n= 679n9 t n= 9n58

Can you find a shorter way to solve this problem?

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Practise

1. Find the next three terms of each arithmetic sequence.

a) 5.8, 7.2, 8.6, b) 34

,54

,74

, ...

2. Given the formula for the nth term of an arithmetic sequence, write the first 4 terms.

a) t n=5 n2 b) t n=n3

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3. Determine which of the following sequences are arithmetic. If a sequence is arithmetic, write the values of a and d.

a) 5, 9, 13, 17,

b) 1, 6, 10, 15, 19,

c) 2, 4, 8, 16, 32,

d) -1, -4, -7, -10,

e) 1, -1, 1, -1, 1,

f) 12

,23 ,

34

,45 ,

g) -4, -2.5, -1, 0.5

h) y , y2 , y3 , y4 ,

k) x , 2 x , 3 x , 4 x ,

l) b , b2 c , b4 c , b6 c ,

4. Given the values of a and d, write the first four terms of each arithmetic sequence.

a) a=52

; d=14 b)

t 1=3 m ; d=1m

5. Find the formula for the nth term for each arithmetic sequence.

a) 9, 16, 23, b) -4, -9, -14, ...

c) 12

, 1 , 32

, 2 , ... d) x , x4, x8 , x12 ,...

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6.- Find the number of terms in each of the following arithmetic sequences.

a) -11, -7, -3, , 153 b) 40, 38, 36, , -30

c) 6, 72 , 1, ... , 104 d) x2, x9, x16, ... , x303

7.- Find a and d, and then write the formula for the nth term, t n , of arithmetic sequences with the following terms.

a) t 2=12 and t 5 = 9 b) t 3= 4 and t 21 =5

8.- The graph of an arithmetic sequence is shown.

a) What are the first terms of the sequence?

b) What is t 50 ? t 200 ?

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Geometric Sequences (Progresiones geomtricas)

In the sequence 2, 10, 50, 250, , each term after the first is found by multiplying the preceding term by 5. Therefore, the ratio of consecutive terms is a constant.

102= 5 ; 50

10= 5 ; 25050 = 5

This type of sequence is called a geometric sequence. The ratio of consecutive terms is

called the common ratio. The first term, t 1 , is denoted by the letter a. The common ratio is denoted by the letter r.

Example 1

Given the formula for the nth term of a sequence, t n=52n1 , write the first 5 terms. Notice that this is a geometric sequence. What is the common ratio for this geometric sequence?

Finding the formula for the n th term of a general geometric sequence

Note that a general geometric sequence where a is the first term and r is the common ratio is:

t 1 = a

t 2 = ar

t 3 = ar r = ar2

t 4 = ar2r = ar

3

...

t n = arn1

where n is a natural number.

Note that r is the ratio of any successive pair of terms. For example,

t2t1=

ar

a= r

t3t2=

a r2

a r= r

Example 2. Finding the Formula for the nth Term.

Find the formula for the nth term, t n , for the geometric sequence 2, 6, 18,

Solution

For the given sequence, a=2 and r=3

t n = a rn1

Substitute known values: t n= 23n1 . The formula for the nth term is t n = 23n1

Now check the three first terms and find t 6 using the formula for the nth term.

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Example 3. Finding the Number of Terms

Find the number of terms in the geometric sequence 3, 6, 12, , 384

Solution

For the given sequence, a=3 , r=2 , and t n = 384Substitute the known values in the formula for the general term and solve for n.

t n= arn1

Substitute known values: 384 = 32n1

Divide both sides by 3: 3843 =

32n1

3Simplify: 128= 2n1

Write 128 as a power of 2: 27 = 2n1

Equate the exponents: 7=n1Solve for n: 8=nThere are 8 terms in the sequence.

Example 4. Finding t n Given Two Terms

In a geometric sequence, t 5= 1875 and t 7= 46875 . Find the first three terms of the sequence and t n

Solution

Substitute known values in the formula for the nth term to write a system of equations. Then, solve the system.

t n= arn1

Write an equation for t 5 : 1875=ar51 1875=ar 4 (1)

Write an equation for t 7 : 46875=ar 71 46875=ar 6 (2)

Divide (2) by (1) : 468751875 =

a r6

a r4 25=r 2

Solve for r : r=5

Since r=5 or r=5 , there are two possible solutions.

A) Substitute 5 for r in (1): B) Substitute - 5 for r in (1):

1875=a54 1875=a54

1875=625 a 1875=625 aSolve for a: 3=a Solve for a: 3=aSince a=3 and r=5 , Since a=3 and r=5 ,the first three terms are 3, 15, and 75. the first three terms are 3, - 15, and 75

The formula for the nth term is t n= 35n1 The formula for the nth term is tn = 35n1

Can you find a shorter way to solve this problem?

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Practise

1.- Determine whether each sequence is arithmetic, geometric, or neither. Then, find the next two terms of each sequence.

a) 1, 4, 9, 16, b) 1, 2, 4, 8,

c) 7, 14, 21, 28, d) 1, 2, 4, 7, 11,

e) 20, 16, 12, 8, f) 32, 16, 8, 4,

g) 113

,103

,83

,53

, ... h) 0.5, 1.5, 4.5, 13.5, ...

2.- State the common ratio and write the next three terms of each geometric sequence.

a) 7, - 7, 7, - 7, b) 0.5, 5, 50, 500, ...

c) 13

,23

,43

,83

, ... d) 800, - 400, 200, - 100, ...

3.- Given the formula for the nth term of a geometric sequence, write the first 4 terms.

a) t n=103n1 b) t n=0.5 4n1

c) t n=23n1

d) t n=1n1

e) t n=2000.5n1 f) t n=10000.1n1

4.- Find the formula for the nth term and find the indicated terms for each of the following geometric sequences.

a) 1, 5, 25, t 6 and t 9

b) 6, 0.6, 0.06, t 6 and t 8

c) 4, - 40, 400, t 8 and t 12

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5.- Find the number of terms in each of the following geometric sequences.

a) 4, 12, 36, , 2916 b) 3, 6, 12, , 1536

c) 4374, 1458, 486, , 2 d) 12

,14

,18

, ... ,1

1024

e) 1

25 ,15 , 1 , ... , 625 f)

281

,4

27,

89 , ... , 6912

6.- Given two terms of each geometric sequence, find t n for the sequence.

a) t 3=36 , t 4=108 b) t 2=6 , t3=12

c) t 2=4 , t 4=64 d) t 3=99 , t 5=11

7.- Motion of a pendulum. On the first swing, a pendulum swings through an arc of 50 cm. On each successive swing, the length of the arc is 0.97 of the previous length. What is the length of the arc, to the nearest hundredth of a centimetre, on

a) the 10th swing?

b) the 15th swing?

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Recursion Formulas (Frmulas de recurrencia)

In previous sections, you have determined sequences using a formula for the nth term. An

example is the formula t n=2n3 which determines the arithmetic sequence 5, 7, 9, 11, 13,

Another example is the formula t n=2n1

which determines the geometric sequence 1, 2, 4, 8,

16,

Such formulas are known as explicit formulas ( frmulas en forma explcita). They can be used to calculate any term in a sequence without knowing the previous term. For example, the tenth

term in the sequence determined by the formula t n=2 n3 is 2(10) + 3, or 23It is sometimes more convenient to calculate a term in a sequence from one or more previous terms in the sequence. Formulas that can be used to do this are called recursion formulas.

A recursion formula consists of at least two parts. The parts give the value(s) of the first term(s) in the sequence, and an equation that can be used to calculate each of the other terms from the term(s) before it.

An example is the formula t 1=5 ; t n=t n12The first part of the formula shows that the first term is 5. The second part of the formula shows that each term after the first term is found by adding 2 to the previous term. The sequence is 5, 7, 9, 11, 13, Thus, this recursion formula determines the same sequence as the explicit

formula tn=2n3 .Practise

Use the recursion formula to write the first 5 terms of each sequence.

a) t 1=3 ; t n=t n12 b) t 1=1 ; t n=2 tn1

c) t 1=6 ; t n=t n12 n d) t 1=2 ; t n=4 t n1n2

e) t 1=3 ; t 2=5 ; t n=t n2t n1 f) t 1=1 ; t 2=2 ; t n=t n2 tn1

g) t 1=1 ; t 2=2 ; t 3=4 ; t n=2 tn3t n23 tn1 h) t 1=1 ; t n1=3tn1

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Investigate

In one of his books, the great Italian mathematician Leonardo de Pisa (Fibonacci) (1180-1250) described the following situation.

A pair of rabbits one month old is too young to produce more rabbits. But at the end of the second month, they produce a pair of rabbits, and a pair of rabbits every month after that. Each new pair of rabbits does the same thing, producing a pair of rabbits every month, starting at the end of the second month.

1.- The table shows how a family of rabbits grows. Extend the table to find the number of pairs of rabbits at the start of the ninth month.

2.- List the numbers of pairs of rabbits in order as the first 9 terms of a sequence. This sequence is known as the Fibonacci sequence. After the first two terms of the sequence, how can each term be calculated from previous terms?

3.- Use the following recursion formula to write the first 9 terms of a sequence. Compare the result with the Fibonacci sequence.

t 1=1 ; t 2=1 ; t n=t n1t n2

4.- Is the Fibonacci sequence arithmetic, geometric, or neither?. Explain

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Arithmetic Series

The following is an arithmetic sequence, 1, 3, 5, 7, 9,

A series is the sum of the terms of a sequence. An arithmetic series is the sum of the terms of an arithmetic sequence.

The series that corresponds to the sequence above is 1 + 3 + 5 + 7 + 9 +

For this series, S5 means the sum of the first 5 terms, so S5=25 .

Investigate

When the mathematician Karl Gauss was eight years old, he used the following method to find the sum of the natural numbers from 1 to 100.

Let S100 represent the sum of the first 100 natural numbers. Write out the series in order and then in reverse order.

S100 = 1 2 3 ... 98 99 100

Reverse: S100 = 100 99 98 ... 3 2 1

Add: 2 S100=101 101 101 ... 101 101 1012 S100=100101

Divide by 2:2S 100

2=

1001012

S100=50101=5050

The same method can be used to derive the formula for the sum of the general arithmetic series. (Suma de n trminos de una progresin aritmtica)

The general arithmetic sequence, a , ad , a2d , ... , t n2d , t nd , t n has n terms, with the first term a and the last term t n .

The corresponding series is Sn = a ad a2 d ... t n2d t nd t nReverse: Sn = t n t nd t n2 d ... a2d ad a

Add: 2 S n = at n at n at n ... at n a tn at n

2 S n=n at n

Divide both sides by 2: Sn =n

2at n

Example 1. Sum of a Series Given First Terms

Find the sum of the first 60 terms of the series, 5 + 8 + 11 +

Solution

We know a = 5 ; d = 3, and n = 60. Formula for the sum of the first n terms: Sn =n

2at n

We need to calculate the last term, t 60

t n = an1d t 60 = 56013 = 5593= 5177= 182

Substitute Sn =n

2at n for the known values S60 =

6025182=30187=5610

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Example 2. Sum of a Series Given First and Last Terms

Find the sum of the arithmetic series, 5 + 9 + 13 + + 201

Solution

Since a = 5 ; d = 4, t n=201 and Sn =n

2at n , we need to calculate the number of

terms, n.

t n = an1d

201 = 5n14 201 = 4n1 200 = 4 n n=50 . There are 50 terms.

Substitute Sn =n

2at n for the known values S50 =

5025201=25206=5150

Practise

1.- Find the sum of

a) the first 100 odd positive integers

b) the first 100 even positive integers

c) the first 75 positive multiples of 3

2.- Find the sum of

a) the positive multiples of 5 up to 500

b) the positive multiples of 7 up to 245

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Geometric Series

The following is a geometric sequence, 3, 6, 12, 24, ...where a = 3 and r = 2

A geometric series is the sum of the terms of a geometric sequence.

The geometric series that corresponds to the sequence above is 3 + 6 + 12 + 24 + ...

For this series, S3 means the sum of the first 3 terms, so S3=3612=21

Investigate

Rosa decided to research her ancestry for the last 6 generations, which included her 2 parents, 4 grandparents, 8 great-grandparents, and so on. To go back 6 generations, the total number of people Rosa needed to research was the sum of the following geometric series.

S6=2481632641.- a) Use addition to find the total number of people Rosa needed to research.

b) To develop a second method for finding the sum of the series, first write the equation that represents the sum of the first 6 terms and label it equation (1).

S6=248163264 (1)Then, multiply both sides of equation (1) by the common ratio, 2, and label the result equation (2).

2 S6=48163264128 (2)Write equations (1) and (2) so that the equal terms on the right side line up as shown.

S6=248163264 (1) 2 S6= 48163264128 (2)

Subtract equation (1) from equation (2). What is the value of S6 ?

2.- a) If S5=1392781 , find S5 by additionb) Predict the next term in the series, and subtract the first term from it.

c) Compare your answer from part b) with your answer from part a).

d) By what factor would you divide your answer from part b) to give the correct sum?

The method developed in the Investigate can be used to write a formula for finding the sum,

Sn , of a general geometric series.

For the general geometric series Sn = a a r a r2 ... a rn1 (1)

Multiply both sides by r: r S n= a r ar2 ... ar

n1 a r

n(2)

Subtract (1) from (2).: r S nSn = a rna

Factor the left side: Sn r1 = a rna

Divide both sides by r1 : Sn =ar

na

r1 or Sn =

a rn1

r1 , r1

So, the sum, Sn , of the n first terms of a geometric series can be found using the formula

Sn =a rn1r1 , r1

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Example 1.- Sum of a Geometric Series When r > 0

Find S8 for the series 2 + 8 + 32 + Solution

a = 2 ; r= 4, and n = 8. Sn =a r

n1

r1

S8 =2 48141

=2 481

3= 43690 . The sum of the first 8 terms is 43690.

Example 2.- Sum of a Geometric Series When r < 0

Find S9 for the series 3 9 + 27 Solution

a = 3 ; r= - 3, and n = 9. Sn =a r

n1

r1

S9 =339131 =

33914

= 14763 . The sum of the first 9 terms is 14763.

Practise

1.- Find the indicated sum for each geometric series.

a) S7 for 1 + 4 + 16 + b) S8 for 2 6 + 18 ...

2.- a) Find the sum S7 for the geometric series 972 + 324 + 108 +

b) Find the sum S15 for the same geometric series 972 + 324 + 108 + If we increase the number of terms what would the sum of all the terms be?

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2.- a) Find the sum S6 for the geometric series 112

14...

b) If we increase the number of terms what would the sum of all the terms be?

3.- Billionaires club. Frank had a plan to become a billionaire. He would put aside 1 cent on the first day, 2 cents on the second day, 4 cents on the third day, and so on, doubling the number of cents each day.

a) How much money would he have after 20 days?

b) How many days would it take Frank to become a billionaire?

c) Do you see any problems with Franks plan? Explain

4.- Chess. According to an old tale, the inventor of chess, Sissa Dahir, was granted anything he wished by the Indian king, Shirham. Sissa asked for one grain of wheat for the first square on the chess board, two grains for the second square, four grains for the third, eight grains for the fourth, and so on, for all 64 squares.

a) How many grains did Sissa ask for?

b) If one grain of wheat has a mass of 65 mg, how did the mass of wheat that Sissa asked for compare with the worlds annual wheat production?

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