Chapter 08 — Sequences and Series — Part 3: Sum of a Finite Arithmetic Series

CHAPTER 08: SEQUENCES AND SERIES

PART 3 — SUM OF A FINITE ARITHMETIC SERIES

Learning objectives

By the end of this lesson, you should be able to:

• Explain the meaning of an arithmetic series.
• Distinguish between an arithmetic sequence and an arithmetic series.
• Derive the formula for the sum of the first n terms of an arithmetic progression.
• Use the sum formula when the first term and common difference are known.
• Use the sum formula when the first and last terms are known.
• Solve problems involving unknown numbers of terms, first terms and common differences.
• Apply arithmetic series to practical situations.

1. Introduction: From sequences to series

In Part 1, we studied arithmetic progressions (APs). An arithmetic progression is a sequence in which the difference between consecutive terms is constant.

Consider the progression:

3, 7, 11, 15, 19

The common difference is 4. The terms form an arithmetic sequence because each term is obtained by adding 4 to the preceding term.

Now suppose we add all these terms:

3 + 7 + 11 + 15 + 19 = 55

This addition is called an arithmetic series.

Important distinction

• Sequence: An ordered list of numbers.
• Series: The sum of the terms of a sequence.

An arithmetic series is therefore the sum of the terms of an arithmetic progression.

2. The sum of the first n terms

Suppose an arithmetic progression has:

• First term = a
• Common difference = d
• Number of terms = n

The progression is:

a, a + d, a + 2d, a + 3d, ..., a + (n − 1)d

We represent the sum of its first n terms by S_n.

S_n = a + (a + d) + (a + 2d) + ... + [a + (n − 1)d]

Our task is to find a convenient formula for this sum without adding every term separately.

3. Derivation of the sum formula from first principles

We begin with the sum written in its normal order:

S_n = a + (a + d) + (a + 2d) + ... + [a + (n − 1)d]   (1)

Now write exactly the same sum in reverse order:

S_n = [a + (n − 1)d] + [a + (n − 2)d] + ... + (a + d) + a   (2)

Add equations (1) and (2) vertically, matching corresponding terms:

First expression	Reverse expression	Pair sum
a	a + (n − 1)d	2a + (n − 1)d
a + d	a + (n − 2)d	2a + (n − 1)d
a + 2d	a + (n − 3)d	2a + (n − 1)d
...	...	...
a + (n − 1)d	a	2a + (n − 1)d

Notice that every pair has the same sum:

2a + (n − 1)d

There are n such pairs. Therefore:

2S_n = n[2a + (n − 1)d]

Divide both sides by 2:

S_n = ⁿ⁄₂[2a + (n − 1)d]

First important formula

S_n = n/2 [2a + (n − 1)d]

This formula gives the sum of the first n terms of an arithmetic progression.

4. An alternative form using the last term

Recall from Part 1 that the nth term of an AP is:

u_n = a + (n − 1)d

Let the last term be represented by l. Then:

l = a + (n − 1)d

Substituting this into the sum formula gives:

S_n = n/2 (a + l)

Second important formula

S_n = n/2 (a + l)

Use this form when the first term, last term and number of terms are known.

Understanding the formula

The formula can be interpreted as:

Sum = Number of terms × Average of first and last terms

The average of the first and last terms is:

(a + l)/2

Multiplying this average by the number of terms gives the total sum.

5. Choosing the correct formula

Information given	Convenient formula
First term a, common difference d, number of terms n	Sn = n/2 [2a + (n − 1)d]
First term a, last term l, number of terms n	Sn = n/2 (a + l)

Remember: Both formulas are equivalent. They produce the same answer when the information is substituted correctly.

6. Worked examples

Example 1: Finding the sum using a and d

Find the sum of the first 10 terms of the arithmetic progression:

3 + 7 + 11 + 15 + ...

Solution

Identify the given information:

• a = 3
• d = 7 − 3 = 4
• n = 10

Use:

S_n = n/2 [2a + (n − 1)d]

Substitute:

S₁₀ = 10/2 [2(3) + (10 − 1)(4)]

= 5[6 + 36]

= 5 × 42

S₁₀ = 210

Answer: 210

Example 2: Finding the sum using the first and last terms

Find the sum of the arithmetic progression:

5 + 9 + 13 + ... + 41

Solution

• a = 5
• l = 41
• d = 4

First find the number of terms using:

l = a + (n − 1)d

41 = 5 + (n − 1)4

36 = 4(n − 1)

9 = n − 1

n = 10

Now use:

S_n = n/2 (a + l)

S₁₀ = 10/2 (5 + 41)

= 5 × 46

S₁₀ = 230

Answer: 230

Example 3: An AP with a negative common difference

Find the sum of the first 12 terms of:

30 + 26 + 22 + 18 + ...

Solution

• a = 30
• d = 26 − 30 = −4
• n = 12

S₁₂ = 12/2 [2(30) + (12 − 1)(−4)]

= 6[60 − 44]

= 6 × 16

S₁₂ = 96

Answer: 96

A negative common difference means that the terms decrease. Be careful to retain the negative sign during substitution.

Example 4: Finding the number of terms

The sum of the first n terms of the AP 4, 7, 10, 13, ... is 175. Find n.

Solution

Given:

• a = 4
• d = 3
• S_n = 175

Use the sum formula:

175 = n/2 [2(4) + (n − 1)3]

175 = n/2 [8 + 3n − 3]

175 = n/2 (3n + 5)

Multiply both sides by 2:

350 = n(3n + 5)

3n² + 5n − 350 = 0

Factorise:

(3n + 35)(n − 10) = 0

Therefore:

n = −35/3   or   n = 10

Since the number of terms must be a positive whole number:

n = 10

Answer: 10 terms.

Example 5: Finding an unknown first term

An arithmetic progression has common difference 3. The sum of its first 8 terms is 116. Find its first term.

Solution

Given:

• d = 3
• n = 8
• S₈ = 116

116 = 8/2 [2a + (8 − 1)3]

116 = 4(2a + 21)

29 = 2a + 21

8 = 2a

a = 4

Answer: The first term is 4.

Example 6: Finding an unknown common difference

The first term of an AP is 6. The sum of its first 10 terms is 240. Find the common difference.

Solution

Given a = 6, n = 10 and S₁₀ = 240.

240 = 10/2 [2(6) + 9d]

240 = 5(12 + 9d)

48 = 12 + 9d

36 = 9d

d = 4

Answer: The common difference is 4.

Example 7: Sum of the first 100 natural numbers

Find the sum:

1 + 2 + 3 + 4 + ... + 100

Solution

This is an AP with:

• a = 1
• d = 1
• n = 100

S₁₀₀ = 100/2 [2(1) + (100 − 1)(1)]

= 50[2 + 99]

= 50 × 101

S₁₀₀ = 5050

Answer: 5050

Example 8: A practical application

A student saves 2,000 shillings in the first week and increases the amount saved by 500 shillings every week. Find the total amount saved over 12 weeks.

Solution

The weekly savings form an AP:

2000, 2500, 3000, 3500, ...

Given:

• a = 2000
• d = 500
• n = 12

S₁₂ = 12/2 [2(2000) + (12 − 1)(500)]

= 6[4000 + 5500]

= 6 × 9500

S₁₂ = 57,000

Answer: The total savings are 57,000 shillings.

Example 9: A sum involving a negative first term

Find the sum of the first 15 terms of:

−8 − 5 − 2 + 1 + ...

Solution

• a = −8
• d = 3
• n = 15

S₁₅ = 15/2 [2(−8) + 14(3)]

= 15/2 [−16 + 42]

= 15/2 × 26

S₁₅ = 195

Answer: 195

Example 10: Finding a sum when the nth term is given

The nth term of an arithmetic progression is given by:

u_n = 5n − 2

Find the sum of the first 20 terms.

Solution

The first term is obtained by putting n = 1:

a = 5(1) − 2 = 3

The second term is:

u₂ = 5(2) − 2 = 8

Therefore d = 8 − 3 = 5.

Now use the sum formula:

S₂₀ = 20/2 [2(3) + 19(5)]

= 10[6 + 95]

= 10 × 101

S₂₀ = 1010

Answer: 1010

7. Finding the sum when the last term is not directly given

Sometimes a question gives the first term, common difference and number of terms. In such cases, the last term can be found first using:

l = a + (n − 1)d

Then use:

S_n = n/2 (a + l)

Alternatively, substitute directly into:

S_n = n/2 [2a + (n − 1)d]

Example 11

Find the sum of the first 25 terms of an AP whose first term is 7 and common difference is 2.

Solution using the last-term method

l = 7 + (25 − 1)2

= 7 + 48 = 55

S₂₅ = 25/2 (7 + 55)

= 25/2 × 62

S₂₅ = 775

Answer: 775

8. Important observations about arithmetic series

Observation 1: Positive common difference

When d is positive, the terms increase. For example:

2 + 5 + 8 + 11 + ...

Observation 2: Negative common difference

When d is negative, the terms decrease. For example:

20 + 16 + 12 + 8 + ...

Observation 3: Zero common difference

When d = 0, all the terms are equal. For example:

5 + 5 + 5 + 5 + 5

The sum of n terms is simply 5n.

Observation 4: The sum can be negative

An arithmetic series may have a negative sum when its terms are sufficiently negative. For example:

−2 − 5 − 8 − 11 = −26

9. Common mistakes to avoid

1. Confusing the nth term with the sum. The nth term is u_n; the sum of the first n terms is S_n.
2. Using the wrong common difference. Always calculate d by subtracting a term from the term immediately after it.
3. Forgetting the negative sign. A decreasing AP has a negative common difference.
4. Using n instead of n − 1. The sum formula contains (n − 1)d.
5. Forgetting to divide by 2. The formula begins with n/2.
6. Using the last-term formula before finding the last term. If l is not supplied, calculate it first or use the formula involving d.
7. Accepting an impossible number of terms. The number of terms must be a positive whole number.

10. Summary of important formulae

1. nth term of an AP

u_n = a + (n − 1)d

2. Sum of the first n terms using the common difference

S_n = n/2 [2a + (n − 1)d]

3. Sum of the first n terms using the last term

S_n = n/2 (a + l)

4. Last term

l = a + (n − 1)d

11. Practice exercises

Attempt all the questions before opening the solutions. Show your working clearly and state the formula used.

Exercise 1

Find the sum of the first 15 terms of the AP:

2 + 5 + 8 + 11 + ...

Show solution

a = 2, d = 3, n = 15.

S₁₅ = 15/2 [4 + 14(3)]

= 15/2 (46)

= 345

Exercise 2

Find the sum of the first 20 terms of:

25 + 22 + 19 + 16 + ...

Show solution

a = 25, d = −3, n = 20.

S₂₀ = 20/2 [50 + 19(−3)]

= 10(50 − 57)

= −70

Exercise 3

Find the sum of the AP:

4 + 9 + 14 + ... + 49

Show solution

a = 4, l = 49, d = 5.

49 = 4 + (n − 1)5

45 = 5(n − 1)

n = 10

S₁₀ = 10/2 (4 + 49)

= 265

Exercise 4

The first term of an AP is 8 and its common difference is 4. Find the sum of the first 18 terms.

Show solution

a = 8, d = 4, n = 18.

S₁₈ = 18/2 [16 + 17(4)]

= 9(84)

= 756

Exercise 5

The sum of the first 12 terms of an AP is 300. If its first term is 3, find the common difference.

Show solution

300 = 12/2 [2(3) + 11d]

300 = 6(6 + 11d)

50 = 6 + 11d

44 = 11d

d = 4

Exercise 6

The sum of the first n terms of the AP 5, 8, 11, 14, ... is 260. Find n.

Show solution

a = 5, d = 3.

260 = n/2 [10 + 3(n − 1)]

520 = n(3n + 7)

3n² + 7n − 520 = 0

(3n + 40)(n − 13) = 0

Since n must be positive:

n = 13

Exercise 7

Find the sum of the first 30 positive even numbers.

Show solution

The AP is 2, 4, 6, 8, ...
a = 2, d = 2, n = 30.

S₃₀ = 30/2 [4 + 29(2)]

= 15(62)

= 930

Exercise 8

An auditorium has 20 rows of seats. The first row has 15 seats, and each successive row has 2 more seats than the previous row. Find the total number of seats.

Show solution

a = 15, d = 2, n = 20.

S₂₀ = 20/2 [30 + 19(2)]

= 10(68)

= 680 seats

Exercise 9

The nth term of an AP is u_n = 4n + 1. Find the sum of the first 25 terms.

Show solution

First term: a = 4(1) + 1 = 5.
Second term = 9, so d = 4.

S₂₅ = 25/2 [10 + 24(4)]

= 25/2 (106)

= 1325

Exercise 10

The first term of an AP is 12 and the tenth term is 39. Find the sum of the first 10 terms.

Show solution

a = 12, l = 39, n = 10.

S₁₀ = 10/2 (12 + 39)

= 5(51)

= 255

Exercise 11

A worker receives 50,000 shillings in the first week and receives an increase of 2,000 shillings each week. Calculate the total amount received over 16 weeks.

Show solution

a = 50,000, d = 2,000, n = 16.

S₁₆ = 16/2 [100,000 + 15(2,000)]

= 8[100,000 + 30,000]

= 8 × 130,000

= 1,040,000 shillings

Exercise 12

The sum of the first 15 terms of an AP is 450. Its first term is 2. Find its common difference and its fifteenth term.

Show solution

Given a = 2, n = 15, S₁₅ = 450.

450 = 15/2 [4 + 14d]

60 = 4 + 14d

56 = 14d

d = 4

Now find the fifteenth term:

u₁₅ = 2 + (15 − 1)4

= 2 + 56

u₁₅ = 58

12. Final lesson check

Before moving to the next part, make sure you can answer these questions:

1. What is the difference between an arithmetic sequence and an arithmetic series?
2. Why do we write the sum in reverse order when deriving the formula?
3. What information is needed to use S_n = n/2 [2a + (n − 1)d]?
4. When is S_n = n/2 (a + l) more convenient?
5. Why must the number of terms be a positive whole number?

Key lesson: The sum of an arithmetic progression can be calculated efficiently by multiplying the number of terms by the average of the first and last terms.

MATHEMATICS HIPHOP

Making Mathematics Easier, One Concept at a Time.

Chapter 08 — Sequences and Series | Part 3

Continue to Part 4: Sum of a Finite Geometric Series.