Patterns and Relationships

A teacher was tired and wanted to rest. To occupy the class, he asked them to add the first 100 natural numbers. The teacher was surprised that in less than no time, one learner had found the answer. This happened at a time when scientific calculators and computers had not been invented. Use a search engine of your choice to find the name of this brilliant learner. Can you find the sum of an arithmetic progression in less than one minute using a short cut? What is the secret?

In this section, you will learn the secret behind this mathematical short cut. First, we will investigate how to find a formula that will help us generate all the terms of both arithmetic and geometric progressions. This formula is usually called the general or nth term of a sequence. Next, you will deduce a formula that will help you to find the sum of the first n terms of linear and exponential sequences.

In finance, sequences and series are used to calculate compound interest as well as interest rates. When designing structures, engineers employ sequences and series.

In computer science, they are used to improve algorithms that facilitate searching and sorting items. In the physical and social sciences, they are used to investigate population growth of humans and bacteria.

KEY IDEAS

• Exponential sequence: If the ratio of any two consecutive terms of a sequence is a constant, it is an exponential sequence or geometric progression.

• Linear sequence: If the difference between any two consecutive terms of a sequence is a constant, it is an arithmetic progression or linear sequence.

• Sequence: A set of numbers in which order is important is known as a sequence. For example, the set of natural numbers is a sequence, or the set of even numbers, or odd numbers are sequences.

• Series: When you replace the commas of a sequence with addition signs, the sequence becomes a series.

Activity 6.1: Investigating natural and artificial patterns In small groups, discuss where you will find patterns on the school campus.

Take your classmates to the place where the patterns are and show the patterns to them. Can you find patterns in nature? Are there natural patterns on campus?

Floor tiles, bathroom tiles and clothing are good examples of patterns created by humans. Games like chess and draught are played on boards with patterns.

The leaves of plants, the teeth and the feathers of some animals have patterns designed by the omnipotent creator, God Almighty.

Activity 6.2: Generating a sequence with equilateral triangles

Figure. 6.1: Triangular numbers Gather some plastic bottle tops and arrange them on a cardboard as shown in

Figure 6.1, noting the number of bottle tops after each step.

How many bottle tops are there in the first arrangement? How many are there in the second arrangement? How many are there in the third and so on? Write the sequence generated on a piece of paper and show it to a classmate. Do you realise that each step creates an equilateral triangle? This is because every side of each triangle has the same number of bottle tops.

In the first arrangement, there is only one bottle top. In the second arrangement, there are three bottle tops. In the third arrangement, there are six. The fourth and fifth arrangements have 10 and 15 bottle tops respectively. Thus, the sequence is 1, 3, 6, 10, 15 … These are known as triangular numbers.

What type of sequence is this? Is it geometric or arithmetic? It is in fact, neither.

There is no constant difference or ratio.

Activity 6.3: Generating a sequence with squares Take a graph paper. Alone or in pairs, carry out, the following activity, step by step.

Step 1: Draw a one-centimetre square on the graph paper and count the number of one-millimetre squares you have in it.

Step 2: Draw a two-centimetre square on the graph paper and count the number of small squares you have in it.

Step 3: Draw a three-centimetre square on the graph paper and count the number of squares you have in it.

Step 4: Continue this until you have a sequence with six terms. Compare your sequence with that of your classmates. Can you predict the number of one-millimetre squares you will find in an eighteen-centimetre square?

What type of sequence is this? Is it linear, exponential, or neither? Explain how you know.

Activity 6.4: Generating a sequence with matchsticks

Step 1: Take a match box. Use three of the matchsticks to form a triangle.

Step 2: Add 3 sticks to one side of this triangle to form a square on one side.

Thus, you have a square attached to the triangle. Note the number of sticks in this arrangement.

Step 3: Add 3 more sticks to the square to form a second square. Note the number of matchsticks in the new arrangement.

Step 4: Add another set of sticks to form a third square and note the number of matchsticks in this arrangement. Write down the sequence generated and show it to a classmate.

Can you predict the number of matchsticks after forming the 20ᵗʰrectangle?

The sequences you have generated so far are man-made sequences. But some sequences occur in nature? For example, the number of new leaves on a germinating plant follow a sequence. This is a sequence known as the Fibonacci sequence and it is found in many places in nature.

Activity 6.5: Generating the sequence of Fibonacci Perform this activity with in pairs or individually.

Step 1: Start with the first two terms of this sequence as 0 and 1.

Step 2: Add the first two terms to get the third term.

Step 3: Add the second and third terms to get the fourth term.

Step 4: Add the third and fourth terms to get the fifth term.

Step 5: Add the fourth and fifth terms to get the sixth term.

Thus, you have the sequence, 0, 1, 1, 2, 3, 5, 8, 13, 21 … These are the first terms of Fibonacci’s sequence. Generate the first 20 terms of Fibonacci’s sequence.

Activity 6.6: Finding the general term of a linear sequence or arithmetic progression

Step 1: Generate a sequence of numbers, if the first term is 3 and the common difference is 5.

This means that, since the first term is 3, add the constant difference, 5, to get the second term. Then add 5 to the second term to get the third term.

Next, add 5 to the third term to get the fourth term and so on. The sequence generated is as follows: 3, 8, 13, 18, 23, 28 …

Step 2: Let the first term of an arithmetic progression or linear sequence be a and let the common difference be d . Write down the first 6 terms of this sequence. Can you predict the 10ᵗʰand 100ᵗʰterms? What will the nth term be?

Table 6.1 displays the terms of the sequence whose first term is a and common difference is d.

Table 6.1: Finding the general term of a linear sequence Position Term Simplified term 1ˢᵗa a + 0d 2ⁿᵈa + d a + 1d 3ʳᵈ(a + d) + d a + 2d 4ᵗʰ(a + d + d) + d a + 3d 5ᵗʰ(a + d + d + d) + d a + 4d ⋮ ⋮ ⋮ 10ᵗʰ(a + d + d + d + d + d + d + d + d) + d a + 9d 100ᵗʰ(a + d + d + d + d + d + d + …) + d a + 99d ⋮ ⋮ ⋮ nᵗʰ(a + d + d + d + d + d + d + …) + d a + d(n − 1) From Table 6.1, you can realise that the general term, Uₙ, is given by Uₙ = a + (n − 1)d.

With this formula, you can determine the position of a given term or the term itself.

Example 6.1

Find the 18ᵗʰand 36ᵗʰterms of the following sequences:

i. 4, 7, 10, …

ii. 13, 7, 6, …

Solution

i. First you have to determine whether the sequence is linear or exponential in order to use the correct formula.

10 − 7 = 7 − 4 = 3. Hence, there is a common difference, d = 3, so it is a linear sequence.

The first term is 4. So a = 4 .

We want the 18ᵗʰterm, so n = 18.

Putting all these values into the formula for the general term:

Uₙ = a + (n − 1)d ⟹ U₁₈ = 4 + (18 − 1)3 = 4 + (17 × 3) = 4 + 51 = 55 Thus, the 18ᵗʰterm of this sequence is 55.

ii. Similarly, the 36ᵗʰterm, U₃₆ = 4 + (36 − 1)3 = 4 + 105 = 109.

Example 6.2

List the set of odd numbers up to the sixth term, starting from one. Find a formula for this sequence. Hence, calculate the 50ᵗʰterm.

Solution

The set of odd numbers, D = {1, 3, 5, 7, 9, 11 …}.

The first term, a = 1 and the common difference, d = 2 .

Consequently, the general term Uₙ = a + (n − 1)d = 1 + 2(n − 1) = 1 + 2n − 2 = 2n − 1.

Hence, the 50ᵗʰterm, U₅₀ = 2(50) − 1 = 100 − 1 = 99.

Example 6.3

Kofi generated a linear sequence such that the 6ᵗʰand 9ᵗʰterms were 19 and 28 respectively. Determine the first term as well as the common difference.

Solution

For any linear sequence, any term is given by the relation Uₙ = a + (n − 1)d.

Thus, U₆ = a + (6 − 1)d = a + 5d. Similarly, the 9ᵗʰterm, U₉ = a + (9 − 1)d = a + 8d.

Thus, you have generated two equations, a + 5d = 19 and a + 8d = 28. Subtracting the first equation from the second, you will get (a + 8d) − (a + 5d) = 28 − 19 ⟹ a + 8d − a − 5d = 9 ⟹ 3d = 9 ∴ d = 3.

Putting d = 3 in a + 5d = 19 or a + 8d = 28 , a + 5(3) = 19 ⟹ a = 19 − 15 = 4.

Thus, the first term is 4 and the common difference is 3.

Now, you are going to unravel the secret behind the mathematical short cut encountered in the introduction: how to find the sum of the first 100 natural numbers in seconds.

Activity 6.7: Finding the sum of the first 10 natural numbers

Step 1: On the first line in your exercise book, write the first 10 natural numbers with an addition sign separating the terms, starting from the lowest to the highest.

Step 2: On the next line in your exercise book, write the first 10 natural numbers with an addition sign separating the terms, this time, starting from the highest to the lowest as shown in the Table 6.2.

Step 3: Add the number on top to the number below it and write the sum beneath each pair of numbers.

Table 6.2: Finding sum of the first 10 natural numbers 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11 Since repeated addition is the same as multiplication, adding 10 elevens is the same as 11 times 10, which is equal to 110. However, by reversing and adding the terms of both sequences, you have doubled the sum. Hence, dividing 110 by 2 gives you the sum of the first 10 numbers. Hence, the sum of the first 10 natural numbers is 55. We can then do a similar task to find the sum of the first 100, or even 1000, natural numbers.

Activity 6.8: Finding the sum of the first 12 even numbers Repeat the steps in activity 6.7, using the first 12 even numbers. See Table 6.3.

Table 6.3: Finding the sum of the first 12 even numbers 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 +24 24 + 22 + 20 + 18 + 16 + 14 + 12 + 10 + 8 + 6 + 4 + 2 26 + 26 + 26 + 26 + 26 + 26 + 26 + 26 + 26 + 26 + 26 + 26 Since repeated addition is the same as multiplication, adding 26 to itself 12 times is the same as 26 times 12, which is equal to 312. However, by reversing and adding the terms of both sequences, you have doubled the sum. Hence, dividing 312 by 2 gives you the sum of the first 12 even numbers. Hence, the sum of the first 12 even numbers is 156.

Activity 6.9: Generating a formula for the sum of the first n terms of a linear sequence To generate a linear sequence whose first term is a and common difference is d , add the common difference to the first term a to get the second term. To get the third term add d to the second term. The fourth term is obtained by adding the common difference to the third term. See table 6.2.

Table 6.4: Formula for the sum of the first n terms Position Terms Terms in reverse order Sum of the terms 1ˢᵗa a + (n − 1)d 2a + (n − 1)d 2ⁿᵈa + d a + (n − 2)d 2a + (n − 1)d 3ʳᵈa + 2d a + (n − 3)d 2a + (n − 1)d 4ᵗʰa + 3d a + (n − 4)d 2a + (n − 1)d 5ᵗʰa + 4d a + (n − 5)d 2a + (n − 1)d 6ᵗʰa + 5d a + (n − 6)d 2a + (n − 1)d 7ᵗʰa + 6d a + (n − 7)d 2a + (n − 1)d 8ᵗʰa + 7d a + (n − 8)d 2a + (n − 1)d ⋮ ⋮ ⋮ ⋮ nᵗʰa + (n − 1)d a 2a + (n − 1)d From the last column of the table 6.2, the sum of each pair of corresponding terms is 2a + (n − 1)d. However, there are n terms. Consequently, the sum of the two sequences is n[2a + (n − 1)d]. Because the reversal doubled the sum, you have to divide by 2 to get Sₙ, which is the sum of the first n terms of any linear sequence.

Thus, the sum of the first n terms of any linear sequence is given by:

Sₙ = n/2[2a + (n − 1)d], where a, d and n are the first term, common difference and position of the term respectively.

Furthermore, Sₙ = n/2[2a + (n − 1)d] = Sₙ = n/2[a + a + (n − 1)d]. But a + (n − 1)d is the general or last term. Hence:

Sₙ = n/2[first term + last term] = n/2[a + l], where l is the last term. This is the secret behind the mathematical trick you encountered in summing the first 100 natural numbers.

Example 6.4

Find the sum of the first 20 terms of the following linear sequences:

i. 6, 9, 12 …

ii. 9, 3, − 3 …

Solution

i. For any linear sequence, the sum of the first n terms is given by:

Sₙ = n/2[2a + (n − 1)d].

d = 12 − 9 = 9 − 6 = 3 = common difference and first term, a = 6.

Therefore, the sum of the first 20 terms, S₂₀, is given by S₂₀ = 20/2 [2(6) + (20 − 1)3] = 10[12 + (19 × 3)] = 10(12 + 57) = 690.

ii. 3 − 9 = − 3 − 3 = − 6 = common difference, d and first term is 9.

Therefore, the sum of the first 20 terms, S₂₀, is given by:

S₂₀ = 20/2 [2(9) + (20 − 1)( − 6)] = 10[18 + (19 × − 6)] = 10(18 − 114) = − 960.

Alternatively, you can use the second formula to find the sum of the first 20 terms of the given sequences.

That is, Sₙ = n/2[first term + last term] = n/2[a + l].

i. There are 20 terms, so the last term U₂₀ = 6 + 3(20 − 1) = 63.

Thus, S₂₀ = 20/2 [6 + 63] = 10(69) = 690.

ii. There are 20 terms, so the last term U₂₀ = 9 + (− 6)(20 − 1) = 9 − 114 = − 105.

Thus, S₂₀ = 20/2 [9 + ( − 105)] = 10(− 96) = − 960.

Example 6.5

The sum of the first n terms of an arithmetic progression is 375.

If the first term is 4 and the common difference is 3, find the number of terms of this sequence.

Solution

The sum of the first n terms of any arithmetic progression is given by Sₙ = n/2 [2a + (n − 1)d].

The first term, a = 4 and the common difference, d = 3. The sum of the first n terms is 375. Thus, you have, Sₙ = n/2[2(4) + (n − 1)3] = 375 ⟹ n/2[8 + 3n − 3] = 375 ⟹ n[3n + 5] = 750 ⟹ 3n²+ 5n − 750 = 0 ⟹ 3n²+ 50n − 45n − 750 = 0 ⟹ n(3n + 50) − 15(3n + 50) = 0 ⟹ (n − 15)(3n + 50) = 0 ∴ n − 15 = 0 or 3 n + 50 = 0 Thus, n = 15 and n = −50/3 .

Since n cannot be negative, n = 15.

That is, the number of terms is 15.

Activity 6.10: Finding the general term, or nth term, of an exponential sequence (or Geometric Progression, or GP)

Step 1: Generate a sequence of numbers, where the first term is 3 and the common ratio is 5. That is, multiply the first term by the common ratio to get the second term. Then multiply the second term by 5 to get the third term and so on. The sequence generated is as follows: 3, 15, 75, 375, 1775 …

Step 2: Let the first term of a geometric progression or exponential sequence be a and let the common ratio be r . Write down the first 5 terms of this sequence. Can you predict the 10ᵗʰand 100ᵗʰterms? What will the nth term be?

In a tabular form, we have:

Table 6.5: Finding nth term of an exponential sequence Position Term Simplified term 1ˢᵗa a r⁰2ⁿᵈa × r a r¹3ʳᵈar × r a r²4ᵗʰa r²× r a r³5ᵗʰa r³× r a r⁴⋮ ⋮ ⋮ 10ᵗʰa r⁸× r a r⁹100ᵗʰa r⁹⁸× r a r⁹⁹⋮ ⋮ ⋮ nᵗʰa rⁿ⁻¹a rⁿ⁻¹From Table 6.5, you can realise that the general term, Uₙ, is given by Uₙ = a rⁿ⁻¹.

With this formula, or relation, you can determine the position of a given term or the term itself in a given exponential sequence.

Example 6.6

Find the 12ᵗʰand 18ᵗʰterms of the following sequences:

i. 1, 2, 4, 8, …

ii. 9, 4.5, 2.25, 1.125, …

Solution

i. First you have to determine whether the sequence is linear or exponential in order to use the correct formula. 8 ÷ 4 = 4 ÷ 2 = 2 ÷ 1 = 2 .

Hence, the common ratio is r = 2 . The first term is 1.

For the 12ᵗʰterm, n = 12.

Substituting all these values into the formula for the general term, you have:

Uₙ = a rⁿ⁻¹⟹ U₁₂ = (1)(2¹²⁻¹) = 2¹¹= 2048.

Thus, the 12ᵗʰterm of this exponential sequence is 2048.

Similarly, the 18ᵗʰterm, U₁₈ = (1)(2¹⁸⁻¹) = 2¹⁷= 131072.

ii. First you have to determine whether the sequence is linear or exponential in order to use the correct formula. 4.5 ÷ 9 = 2.25 ÷ 4.5 = 1.125 ÷ 2.25 = 1/2.

Hence, the common ratio is r = 1/2. The first term is 9.

For the 12ᵗʰterm, n = 12.

Substituting all these values into the formula for the general term, you have:

Uₙ= arⁿ⁻¹⟹ U₁₂= (9)(0.5¹²⁻¹) = 9/2048 Thus, the 12ᵗʰterm of this exponential sequence is 9/2048.

Similarly, the 18ᵗʰterm, U₁₈= (9)(0.5¹⁸⁻¹) = 9/131072.

Example 6.7

Consider the following sequence: 1/32 , 1/16 ,1/8 … , 32.

a. Find a general formula for this sequence.

b. How many terms are there in this sequence?

Solution

This is particularly tricky with a lot of indices work, so go through this in a small group, explaining each step to each other.

First, you have to determine the type of sequence in order to use the correct formula.

Since 1/8 ÷ 1/16 = 1/16 ÷ 1/32 = 2 , the common ratio is 2. Therefore, this is an exponential sequence.

The first term is 1/32. Putting these values into the equation we have:

Uₙ = a rⁿ⁻¹= ( 1_

32) 2ⁿ⁻¹= ( 1/2⁵)(2ⁿ)(2⁻¹) = (2⁻⁵)(2ⁿ)(2⁻¹) = (2⁻⁵⁻¹)(2ⁿ) = (2⁻⁶)(2ⁿ) = 2−6 + n = 2ⁿ− 6 Thus, the general term, Uₙ of the sequence is given by Uₙ = 2ⁿ– 6 = 2ⁿ__ 2⁶= 2ⁿ__ 64 Since, the last term is 32 and the general term can be used to find any term, you have the equation 2ⁿ__ 64 = 32.

Now, 2ⁿ__ 64 = 32 ⟹ 2ⁿ= 32 × 64 = 2⁵× 2⁶= 2⁵⁺⁶= 2¹¹ ⟹ 2ⁿ= 2¹¹n = 11 Thus, there are 11 terms in this sequence.

Example 6.8

The 6ᵗʰand 9ᵗʰterms of a geometric progression are 3 and 81 respectively.

Determine the first term as well as the common ratio.

Solution

For any exponential sequence, any term is given by the relation Uₙ = a rⁿ⁻¹.

Thus, U₆ = a r⁶⁻¹= a r⁵= 3. Similarly, the 9ᵗʰterm, U₉ = a r⁸= 81.

Now, U₉__ U₆ = a r⁸___ a r⁵= 81/3 a r⁸_ a r⁵= 81/3 r⁸_ r⁵= 27 r⁸⁻⁵= 3³r³= 3³r = 3 Now, put r = 3 in the equation a r⁵= 3 , a (3⁵) = 3 ⟹ 243a = 3 ⟹ a = 3/243 = 1/81.

Thus, the first term is 1/81 and the common ratio is 3.

Activity 6.11: Deriving the formula for the sum of the first n terms of geometric progression.

Step 1:

Generate an exponential sequence in which the first term is a and the common ratio is r .

a, ar, ar², ar³, a r⁴, a r⁵, … , a rⁿ⁻², a rⁿ⁻¹.

When the terms of this sequence are added, you will get the sum of the first n terms of an exponential sequence.

Thus, Sₙ = a + ar+ ar²+ ar³+ a r⁴+ a r⁵+ … + a rⁿ⁻²+ a rⁿ⁻¹.

The sum of the terms of a sequence is also known as a series.

Let us multiply the series Sₙ by r :

SSnn = aa + aaaa + aaaa²²+ aaaa³³+ aaaa⁴⁴+ aaaa⁵⁵+ … + aaaann&22 + aaaann&11 ⟹ aaSSnn = aaaa + aaaa²²+ aaaa³³+ aaaa⁴⁴+ aaaa⁵⁵+ … + aaaann&22 + aaaann&11 + aaaaⁿⁿ.

By now you will have realised that the only difference between the series Sₙ and r Sₙ below it are the first term a and the last term a rⁿ. Thus, when you subtract Sₙ from r Sₙ or vice versa, you will get Sₙ − r Sₙ = a − a rⁿor r Sₙ− Sₙ = a rⁿ− a .

Thus, Sₙ(1 − r) = a(1− rⁿ).

Thus, the sum of the first n terms of an exponential sequence, Sₙ is given by:

Sₙ = a(1− rⁿ)______ 1 − r , we tend to use this if r < 1 and Sₙ = a(rⁿ− 1)______ r − 1 , and use this if r > 1.

But either work all the time.

Example 6.9

Find the sum of the first 12 terms of the following geometric progressions or exponential sequence:

i. 6, 3, 1.5, 0.75 …

ii. 1/27 , 1/9 , 1/3 …

Solution

i. For any exponential sequence, the sum of the first n terms is given by:

Sₙ = a(1− rⁿ)______ 1 − r , where r < 1 and Sₙ = a(rⁿ− 1)______ r − 1 , where r > 1.

Now,1.5 ÷ 3 = 3 ÷ 6 = 0.5 =common ratio,r . Sincer < 1 , we will use the formula Sₙ = a(1− rⁿ)______ 1 − r , where r < 1 (but the other formula works too!)

Now, the first term a = 6 and n = 12 since you want to find the sum of the first 12 terms.

Thus, S₁₂ = 6(1− 0.5¹²)________ 1 − 0.5 = 12(1− 0.5¹²) = 11.997 (3 decimal places).

ii. 1/9 ÷ 1/27 = 1/3 ÷ 1/9 = 3 = common ratio. Since r , the common ratio, is 3 > 1 , we will use the formula Sₙ = a(rⁿ− 1)______ r − 1 (but the other one would work too).

Therefore, S₁₂ = 1/27(3¹²− 1) ________ 3 − 1 = 3¹²− 1/54 = 531441 − 1/54 = 531440/54 = 9841.48 (to 2 decimal places).

If the common ratio, r < 1 , then you can find the sum of the series as the number of terms n approaches infinity. The sum of an infinite series as n approaches infinity (n → ∞) is given by the formula S∞= a/1− r.

Example 6.10

Find the sum of the geometric progression, 1, 1/3, 1/9, 1/27 , … as the number of terms approaches infinity.

Solution

The common ratio, r = 1/3 and the first term a = 1.

Thus, the sum to infinity, S∞ = 1/1− 1/3 = 1__ 2/3 = 3/2 = 11/2.

We will now look at how the concept of sequences and series are applied in various fields.

Banking: Compound interest When an amount of money is deposited at a bank, the bank owes the customer interest because the bank uses the customer’s money to make a profit. The interest on the deposited amount depends on the interest rate of the bank. If the bank’s rate of interest is 10% and you deposit GH¢100.00 on the first of January 2025, on the first of January, 2026, GH¢10.00 will be deposited into your account as the interest. Thus, the amount in your account will be GH¢110.00. Table 6.6 will give you the amount of money in the account at the beginning of every year.

Table 6.6: Calculating compound interest Year Amount Simplified amount Amount 1 100 100 × (110_ 100) 0 GH¢100.00 2 100 × 110/100 100 × (110_ 100) 1 GH¢110.00 Year Amount Simplified amount Amount 3 100 × 110/100 × 110/100 100 × (110_ 100) 2 GH¢121.00 4 100 × 110/100 × 110/100 × 110/100 100 × (110_ 100) 3 GH¢133.10 5 100 × 110/100 × 110/100 × 110/100 × 110/100 100 × (110_ 100) 4 GH¢146.41 ⋮ ⋮ ⋮ nth 100 × (110_ 100) n−1 GH¢100 (1.1)ⁿ⁻¹Thus, at the beginning of the nᵗʰyear the amount Uₙ = 100 × (110/100) n−1 = 100 (1.1)ⁿ⁻¹.

By now, you will have noticed that the sequence generated in table 6.4 is a geometric progression whose first term, a = 100 and the constant ratio, r = 110/100 = 1.1. In this way, sequences and series can be used to calculate the compound interest for customers who have deposits at the bank. Thus, you can use the concept of sequences and series to calculate the compound interest without generating the sequence.

Example 6.9

Mr. Mann deposited GH¢1800.00 in a bank whose interest rate was 9% per annum.

Calculate the money in his account after the:

a. 3ʳᵈyear;

b. 6ᵗʰyear;

c. 12ᵗʰyear.

Solution

At the end of the nᵗʰyear, the amount of money, Uₙ, in his account is given by Uₙ = a rⁿ.

Thus, Uₙ = 1800 (100 + 9/100 ) n = 1800 (109/100) n = 1800 (1.09)ⁿa. U₃ = 1800 (1.09)³= 2331.0522 ≈ GH¢2331.05

b. U₆ = 1800 (1.09)⁶= 3018.7802 ≈ GH¢3018.78

c. U₁₂ = 1800 (1.09)¹²= 5062.7966 ≈ GH¢5062.80

Example 6.10

When Mr. Ackah started his teaching career his monthly salary was GH¢500.00.

It was agreed that every year GH¢100.00 will be added to his salary.

a. Calculate the salary earned after 10 years in the teaching service.

b. How long will it take him to earn GH¢5000.00 a month?

Solution

Forming the sequence for this situation, you will have: 500, 600, 700, … Thus, you have a linear sequence in which the first term, a = 500 and common difference, d = 100 .

a. The general term, Uₙ = a + (n − 1)d and the salary after 10 years means that the n = 10.

∴ U₁₀ = 500 + (10 − 1)(100) = 500 + 900 = 1400.

Thus, after 10 years of teaching his salary was GH¢1400.00.

b. Now, the last or general term, Uₙ = 5000.

Thus, 500 + (n − 1)100 = 5000 ⟹ (n − 1)100 = 5000 − 500 = 4500 ⟹ n − 1 = 45 ⟹ n = 46 Thus, it will take him 46 years to earn a salary of GH¢5000.00 per month

Example 6.11

Madam Anvo decided to save 12% of her monthly salary every month for her pension. If her salary was GH¢8000.00, calculate the amount saved for her pension after 15 years.

Solution

First, you have to find the amount saved for her pension, which is 12% of the salary.

Thus, the amount saved per month is 12/100 × 8000 = 0.12 × 8000 = GH¢960.00 .

This amount can be used to form a sequence: 960, 1920, 2880, 3840, … Obviously, this is a linear sequence, where the first term is 960 and the common difference is 960.

For this reason, U₁₅ = 960 + (15 − 1)960 = 960 + 14(960) = 960 + 13440 = GH¢14 400.00.

Consequently, after 15 years, Madam Anvo would have saved GH¢14 400.00.

Social sciences In the social sciences like geography and economics, sequences and series are used to anticipate or predict the population of regions or countries. This helps social scientists and politicians to plan for the future.

Example 6.12

The population growth of Africa’s most populous nation, Nigeria, is approximately 3% per year. If the population of Nigeria in 2024 is 240 million, calculate the population of Nigeria in 2030 to the nearest million.

Solution

The population growth will obey an exponential sequence.

The first term a = 240 000 000 and the common ratio r = 100 + 3/100 = 103/100 = 1.03.

Thus, the general term, Pₙ, for the yearly population of Nigeria is given by Pₙ = a rⁿ= 240 000 000 (1.03)ⁿ.

Since the difference between 2024 and 2030 is 6 years, the value of n is 6.

Thus, P₆ = 240, 000, 000 (1.03)⁶= 286 572 551.16696 ≅ 287million (to the nearest million) Physical sciences In the physical sciences like biology, it can be important to determine the rate at which living things grow. The chemist might also be interested in the rate at which a radioactive substance decay. The botanist may be interested in the rate at which a particular tree grows. In all such situations, this cannot be done without sequences and series.

Example 6.13

A biochemist realised that the growth rate of bacteria in a medium is 5% per day.

If there were about 1200 bacteria at the beginning of the experiment, how many bacteria were there after 10 days?

Solution

The population growth of the bacteria in the medium follows an exponential sequence in which the first term is 1200 and the common ratio is 1.05.

That is, a = 1200 , r = 1.05 , n = 10.

Putting all these values into the relation, Pₙ = a rⁿ= 1200 (1.05)ⁿ.

⟹ P₆ = 1200 (1.05)⁶= 1608.11476875. Thus, there will be approximately 1 600 bacteria in the medium after 10 days.

Example 6.14

Researchers sent out by the Forestry Commission calculated that a particular tree grows by 0.9 metres every year.

If the tree was 2.7 metres tall this year, how long will it take the tree to reach a height of 36 metres?

Solution

You can generate the sequence to know the type of sequence you are dealing with.

2.7, (2.7 + 0.9), (2.7 + 0.9 + 0.9), (2.7 + 0.9 + 0.9 + 0.9), … = 2.7, 3.6, 4.5, 5.4, … It is clear that the sequence is linear with the first term, a = 2.7 and the common difference is 0.9.

Thus, you have to use the relation Uₙ = a + (n − 1)d.

⟹ 2.7 + (n − 1)0.9 = 36 ⟹ (n − 1)0.9 = 36 − 2.7 = 33.3 n − 1 = 33.3/0.9 = 37 ∴ n = 37 + 1 = 38 Thus, it will take the tree 38 years to reach a height of 36 metres.

Social Services

Hotels and supermarkets sometimes display items in attractive geometric shapes to attract customers. In such a case, there is the need to apply sequences and series to calculate the number of tins in a display.

Example 6.15

Yaba displays the tins of milo in her shop to form a pyramid. There are 15 tins in the first or bottom row of this pyramid and just one tin on the topmost row. If each successive row has 2 fewer tins than the row below it, calculate the total number of milo tins in the display.

Solution

The sequence in the question is given by 15, 13, 11, 9… 1.

Thus, the first term, a = 15 and the common difference, d = 2 . The total number of tins in the display is the same as the sum of the terms of this sequence. You will notice that this is a linear series in which the constant difference is -2 and n = 8 .

Thus, the sum, Sₙ = n/2(a + l), where a, l and n are the first term, last term and number of terms respectively.

∴ S₈ = 8_ 2(15 + 1) = 4(16) = 64. This means that there are 64 tins in the display.

Auditing Auditors may be interested in the value at which an institution disposed of its worn-out assets. In such a case, calculating the rate of depreciation of a vehicle or machine can help them to know whether the institution got value for money.

Sequences and series are important tools when calculating the rate of depreciation of an item.

Example 6.16

The auditors calculated that the school bus depreciates by 9% of its initial value every year. If the value of the bus was GH¢ 144 000.00 in 2023, calculate the value of the bus in 2028?

Solution

Depreciation of items normally follows a geometric progression in which the first term is the initial value of the bus (cost price) and the rate of depreciation is 12%.

Thus, the common ratio is 100 % − 12 % = 0.88 = r . Therefore, you have to use the general term of a geometric progression, Uₙ = a rⁿ, where n = 2028 − 2023 = 5.

U₅ = 144 000 × 0.88⁵= 75993. 396 ≈ GH¢75, 993.40 Thus, in 2028, the value of the bus will be approximately GH¢75 993.40.

1. The fifth term of an arithmetic progression is 18 and the eleventh term is

36. Find the sum of the first term and the common difference.

A. 9 B. 8 C. 7 D. 6

2. Find the sum of the odd numbers from 1 up to 99.

A. 5000 B. 2500 C. 2050 D. 1050

3. If 1, 2x + 1, 5, 3x + 4, 9, … is a linear sequence, find the:

a. value of the constant x ;

b. first 8 terms of the sequence;

c. 15ᵗʰterm;

d. sum of the first 21 terms of the sequence.

4. The sum of the first 6 terms of an arithmetic progression is 21 and the sum of the first 10 terms is 55. Find the sum of the first twenty terms of this sequence.

5. Determine whether the following sets of numbers form linear or exponential sequences.

i. 1/2, 1, 3/2 , 2, …

ii. 1/2 , 1/3 , 1/4 , 1/5 , …

iii. 3, 1, 1/3 , 1/9 , …

6. Write down the first six terms of the geometric progression in which the first term is 2 and the common ratio is 1/3.

7. Find the general term of the geometric progression whose first term is 6 and common ratio is 3. Hence, find the 10ᵗʰand 20ᵗʰterms of the progression.

8. Find the ninth term of the exponential sequence 3, − 6, 12, . . .

9. Find the sum of the first eight terms of a geometric progression, if the first term is 2 and common ratio is 3.

10. If the first term and common ratio of a GP are 4 and 9/10 respectively, find the number of terms such that the sum of the sequence is more than 36.

11. a. Draw a table of values for the relation y = 3ˣfor the interval 0 ≤ × ≤ 5.

b. Using a scale of 2cm to 50 units on the y -axis and 2cm to 1 unit on the x -axis, draw the graph of the relation y = 3ˣwithin the given interval.

c. What type of sequence do the y-values form?

d. What conclusion can you draw about the graph of an exponential sequence?

12. a. Draw a table of values for the relation y = 3x − 1 for the interval 0 ≤ × ≤ 5.

b. Using a scale of 2cm to 2 units on the y -axis and 2cm to 1 unit on the x -axis, draw the graph of the relation y = 3x − 1 within the given interval.

c. What type of sequence do the y-values form?

d. What conclusion can you draw about the graph of an arithmetic progression?