Arithmetic Progressions

Defining Sequences and Arithmetic Progressions

In Maths, a sequence is a list of numbers written in a special order. Each number in the list is called a term. Sequences can follow many different patterns.

For example, the sequence \(2, 4, 6, 8, 10\) increases by 2 each time. The sequence \(1, 4, 9, 16\) does not increase by the same amount each time, even though it still follows a pattern.

In this lesson, you will learn how to tell whether a sequence is just a general pattern or a special kind of sequence called an arithmetic progression (often shortened to AP).

1. What is a sequence?

A sequence is an ordered list of numbers. Order matters. This means \(3, 5, 7, 9\) is a different sequence from \(9, 7, 5, 3\).

We often label terms using position numbers:

• 1st term: \(a_1\)
• 2nd term: \(a_2\)
• 3rd term: \(a_3\)
• nth term: \(a_n\)

The symbol \(a_n\) means “the term in position \(n\).”

2. What is an arithmetic progression?

An arithmetic progression is a sequence where the difference between consecutive terms is always the same.

This fixed amount is called the common difference, written as \(d\).

For example, in the sequence \(5, 8, 11, 14, 17\):

• \(8 - 5 = 3\)
• \(11 - 8 = 3\)
• \(14 - 11 = 3\)
• \(17 - 14 = 3\)

Because the difference is always \(3\), this sequence is an arithmetic progression.

3. How to check whether a sequence is arithmetic

To decide if a sequence is an arithmetic progression, subtract each term from the term after it.

If all the differences are equal, then the sequence is arithmetic. If the differences are not all equal, then it is not an arithmetic progression.

Here is the process:

1. Write the sequence clearly.
2. Find the difference between consecutive terms.
3. Compare the differences.
4. If they are all the same, identify the common difference \(d\).

Example: Check whether \(12, 9, 6, 3, 0\) is arithmetic.

• \(9 - 12 = -3\)
• \(6 - 9 = -3\)
• \(3 - 6 = -3\)
• \(0 - 3 = -3\)

All the differences are \(-3\), so this is an arithmetic progression with common difference \(d = -3\).

Important: The common difference can be positive, negative, or zero.

• Positive difference: the terms increase
• Negative difference: the terms decrease
• Zero difference: all terms stay the same

For instance, \(7, 7, 7, 7\) is still an arithmetic progression because the common difference is \(0\).

4. General sequences versus arithmetic progressions

Not every sequence is arithmetic. Some sequences follow other types of patterns.

Look at these examples:

• \(3, 6, 9, 12, 15\) → differences are \(+3, +3, +3, +3\), so this is arithmetic.
• \(2, 4, 8, 16\) → differences are \(+2, +4, +8\), so this is not arithmetic.
• \(1, 4, 9, 16\) → differences are \(+3, +5, +7\), so this is not arithmetic.
• \(20, 15, 10, 5, 0\) → differences are \(-5, -5, -5, -5\), so this is arithmetic.

A sequence can have a pattern without being arithmetic. For it to be an arithmetic progression, the pattern must involve adding or subtracting the same number each time.

5. Parts of an arithmetic progression

An arithmetic progression is usually described using:

• First term: \(a_1\)
• Common difference: \(d\)
• nth term: \(a_n\)

If you know the first term and the common difference, you can build the whole sequence.

For example, if \(a_1 = 4\) and \(d = 6\), the sequence is:

\(4, 10, 16, 22, 28, \dots\)

Each new term is found by adding \(6\).

6. The nth term formula for an arithmetic progression

The nth term of an arithmetic progression can be found using this formula:

This formula works because:

• The first term is \(a_1\)
• To get to the second term, you add \(d\) once
• To get to the third term, you add \(d\) twice
• To get to the nth term, you add \(d\) a total of \(n-1\) times

This formula helps you find any term without writing the entire sequence.

Worked Example 1: Decide whether a sequence is arithmetic

Determine whether \(7, 11, 15, 19, 23\) is an arithmetic progression.

Step 1: Find the differences.

• \(11 - 7 = 4\)
• \(15 - 11 = 4\)
• \(19 - 15 = 4\)
• \(23 - 19 = 4\)

Step 2: Compare the differences.

They are all equal to \(4\).

Conclusion: Yes, this is an arithmetic progression with common difference \(d = 4\).

Worked Example 2: Decide whether a sequence is not arithmetic

Determine whether \(2, 5, 10, 17, 26\) is an arithmetic progression.

Step 1: Find the differences.

• \(5 - 2 = 3\)
• \(10 - 5 = 5\)
• \(17 - 10 = 7\)
• \(26 - 17 = 9\)

Step 2: Compare the differences.

The differences are \(3, 5, 7, 9\), which are not all the same.

Conclusion: This sequence is not an arithmetic progression.

Worked Example 3: Find the nth term

Find the nth term of the arithmetic progression \(6, 10, 14, 18, 22, \dots\)

Step 1: Identify the first term and common difference.

• First term: \(a_1 = 6\)
• Common difference: \(d = 4\)

Step 2: Use the formula.

Substitute the values:

Simplify:

Answer: The nth term is

You can check it:

• When \(n=1\), \(a_1 = 4(1)+2 = 6\)
• When \(n=2\), \(a_2 = 4(2)+2 = 10\)

Worked Example 4: Find a specific term

The first term of an arithmetic progression is \(3\), and the common difference is \(5\). Find the 12th term.

Step 1: Write the formula.

Step 2: Substitute \(a_1 = 3\), \(d = 5\), and \(n = 12\).

Answer: The 12th term is \(58\).

7. Writing an arithmetic progression from given information

Sometimes you are given the first term and the common difference, and you must write the sequence.

Example: Write the first five terms of an arithmetic progression with \(a_1 = -2\) and \(d = 3\).

Start with \(-2\), then keep adding \(3\):

• 1st term: \(-2\)
• 2nd term: \(1\)
• 3rd term: \(4\)
• 4th term: \(7\)
• 5th term: \(10\)

So the sequence is \(-2, 1, 4, 7, 10\).

8. Common mistakes to avoid

• Looking only at the first two terms: You must check all consecutive differences given.
• Mixing up subtraction order: Always do “next term minus previous term.”
• Assuming every pattern is arithmetic: A pattern may exist without a constant difference.
• Using the nth term formula on a non-arithmetic sequence: First verify the common difference is constant.

9. Quick test for yourself

• Is \(14, 18, 22, 26\) arithmetic? Yes, because the difference is always \(4\).
• Is \(5, 10, 20, 40\) arithmetic? No, because the differences are \(5, 10, 20\).
• If \(a_1 = 9\) and \(d = -2\), what are the first four terms? \(9, 7, 5, 3\).

Brief Summary

A sequence is an ordered list of numbers, and an arithmetic progression is a special sequence with a constant common difference.

To check if a sequence is arithmetic, subtract consecutive terms and see whether the difference stays the same. If it does, the sequence is arithmetic.

For an arithmetic progression, the nth term formula is

This formula lets you describe the pattern and find any term in the sequence.

Lesson: The General Term of an Arithmetic Progression

In many number patterns, the same amount is added each time to get the next term. These patterns are called arithmetic progressions, or APs.

For example, in the sequence 3, 7, 11, 15, 19, ... each term increases by 4. Because the increase is constant, this is an arithmetic progression.

In this lesson, you will learn how to find a rule for any term in an arithmetic progression. This rule is called the general term, and it helps you find any term without listing all the earlier ones.

1. What is an arithmetic progression?

An arithmetic progression is a sequence in which the difference between consecutive terms is always the same.

• The first term is usually called \(a\).
• The common difference is usually called \(d\).
• The nth term, or general term, is written as \(a_n\).

For example, in the sequence 10, 13, 16, 19, 22, ...

• First term: \(a = 10\)
• Common difference: \(d = 3\)

We can see the pattern by writing each term as repeated addition:

\(a_1 = 10\)
\(a_2 = 10 + 3\)
\(a_3 = 10 + 3 + 3 = 10 + 2(3)\)
\(a_4 = 10 + 3 + 3 + 3 = 10 + 3(3)\)

This shows an important pattern: to get the \(n\)th term, we start with the first term and add the common difference \(n-1\) times.

2. The formula for the general term

The general term of an arithmetic progression is:

This formula means:

• Start with the first term \(a\)
• Multiply the common difference \(d\) by \(n-1\)
• Add the result to \(a\)

Why is it \(n-1\) and not \(n\)?

Because the first term already starts at \(a\). To reach the 2nd term, you add the difference once. To reach the 3rd term, you add it twice. So to reach the \(n\)th term, you add the difference \(n-1\) times.

3. Understanding the formula with a pattern

Suppose the arithmetic progression is:

5, 8, 11, 14, 17, ...

Here:

• \(a = 5\)
• \(d = 3\)

Using the formula:

So the general term is:

This can also be simplified:

So both of these are correct:

• \(a_n = 5 + 3(n-1)\)
• \(a_n = 3n + 2\)

They mean the same thing.

4. How to find the general term

When you are given an arithmetic progression, follow these steps:

1. Find the first term \(a\).
2. Find the common difference \(d\) by subtracting one term from the next.
3. Substitute into the formula \(a_n = a + (n-1)d\).
4. Simplify if needed.

Worked Example 1: Find the general term

Find the general term of the arithmetic progression:

4, 9, 14, 19, 24, ...

Step 1: Identify the first term.

\(a = 4\)

Step 2: Find the common difference.

\(9 - 4 = 5\), so \(d = 5\)

Step 3: Use the formula.

Step 4: Simplify.

Answer: The general term is \(a_n = 5n - 1\).

Check:

• For \(n=1\): \(5(1)-1=4\)
• For \(n=2\): \(5(2)-1=9\)
• For \(n=3\): \(5(3)-1=14\)

The rule works.

Worked Example 2: Find a specific term

The arithmetic progression is:

12, 17, 22, 27, ...

Find the 20th term.

Step 1: Identify \(a\) and \(d\).

• \(a = 12\)
• \(d = 5\)

Step 2: Use the formula with \(n = 20\).

Answer: The 20th term is \(107\).

This shows why the formula is useful. You do not need to write all 20 terms to find the answer.

Worked Example 3: Arithmetic progression with a negative difference

Find the general term of the sequence:

30, 26, 22, 18, 14, ...

This is still an arithmetic progression because the change is constant.

Step 1: First term:

\(a = 30\)

Step 2: Common difference:

\(26 - 30 = -4\), so \(d = -4\)

Step 3: Use the formula.

Step 4: Simplify.

Answer: The general term is \(a_n = 34 - 4n\).

Check: For \(n=1\), \(34 - 4 = 30\), which matches the first term.

Worked Example 4: Find which term has a given value

The arithmetic progression is:

7, 11, 15, 19, 23, ...

Which term is equal to 63?

Step 1: Identify \(a\) and \(d\).

• \(a = 7\)
• \(d = 4\)

Step 2: Write the general term.

Step 3: Set \(a_n = 63\).

Step 4: Solve.

Answer: 63 is the 15th term.

5. Common mistakes to avoid

• Using \(n\) instead of \(n-1\): The correct formula is \(a_n = a + (n-1)d\).
• Choosing the wrong common difference: Always subtract one term from the next carefully.
• Forgetting negative differences: If the sequence decreases, \(d\) is negative.
• Mixing up term number and term value: In \(a_{10}\), the 10 tells you the position, not the value.

6. Quick practice ideas

Try these on your own:

• Find the general term of 2, 6, 10, 14, ...
• Find the 12th term of 15, 12, 9, 6, ...
• In the sequence 9, 13, 17, 21, ..., which term is 49?

7. Summary

An arithmetic progression has a constant difference between terms. If the first term is \(a\) and the common difference is \(d\), then the general term is:

This formula helps you:

• find the rule for the sequence,
• calculate any term directly,
• work out which term has a certain value.

Whenever you see a sequence going up or down by the same amount each time, think of the arithmetic progression formula.

Finding Specific Terms from the End of an AP

An Arithmetic Progression (AP) is a sequence of numbers in which the difference between consecutive terms is always the same. This fixed difference is called the common difference.

For example, in the AP 3, 7, 11, 15, 19, the common difference is 4 because each term increases by 4.

Usually, when we work with an AP, we count terms from the beginning: 1st term, 2nd term, 3rd term, and so on. But sometimes a question asks for a term from the end, such as the 2nd term from the end or the 5th term from the end.

In a finite AP, this means we count backward from the last term. Learning how to do this correctly is very useful in many problems.

1. Recall the basic formula for the nth term of an AP

If an AP has first term \(a\) and common difference \(d\), then the nth term is:

$$T_n = a + (n-1)d$$

Here:

• \(a\) = first term
• \(d\) = common difference
• \(n\) = position of the term

If the AP has a last term, we often call it \(l\).

So, if there are \(n\) terms in the AP, then:

$$l = a + (n-1)d$$

2. What does “kth term from the end” mean?

The 1st term from the end is the last term.

The 2nd term from the end is the term just before the last term.

The 3rd term from the end is two places before the last term, and so on.

So if the last term is \(l\), then:

• 1st term from the end = \(l\)
• 2nd term from the end = \(l-d\)
• 3rd term from the end = \(l-2d\)
• 4th term from the end = \(l-3d\)

This creates a pattern.

Therefore, the kth term from the end is:

$$l - (k-1)d$$

This is the most important formula for this topic.

3. Formula for the kth term from the end

If the last term of a finite AP is \(l\), then:

$$\text{kth term from the end} = l - (k-1)d$$

This formula works because every step backward in an AP changes the term by \(d\).

4. Another way to think about it

Suppose an AP has \(n\) terms. Then:

• 1st term from the end = \(n\)th term
• 2nd term from the end = \((n-1)\)th term
• 3rd term from the end = \((n-2)\)th term

So the kth term from the end is really the:

$$\big(n-k+1\big)\text{th term from the beginning}$$

Then we can use:

$$T_{n-k+1} = a + (n-k)d$$

Since \(l = a + (n-1)d\), this gives the same result:

$$l - (k-1)d$$

5. Steps to find a term from the end

1. Identify the common difference \(d\).
2. Find the last term \(l\), if it is given or can be found.
3. Identify which term from the end is needed, say the \(k\)th term.
4. Use the formula:
   $$\text{kth term from the end} = l - (k-1)d$$

Be careful with the value of \(k\). For example:

• 2nd from the end means subtract \(d\) once.
• 5th from the end means subtract \(d\) four times.

Worked Example 1: Finding the 3rd term from the end

Find the 3rd term from the end of the AP:

$$5, 9, 13, 17, 21, 25$$

Step 1: Identify the last term.

Here, \(l = 25\).

Step 2: Find the common difference.

Each term increases by 4, so \(d = 4\).

Step 3: Use the formula for the 3rd term from the end.

Here, \(k = 3\).

$$\text{3rd term from the end} = l - (3-1)d$$

$$= 25 - 2(4)$$

$$= 25 - 8 = 17$$

Answer: The 3rd term from the end is 17.

You can check by counting backward: 25, 21, 17.

Worked Example 2: Finding the 5th term from the end

Find the 5th term from the end of the AP:

$$2, 5, 8, 11, 14, 17, 20, 23, 26$$

Step 1: Last term is \(l = 26\).

Step 2: Common difference is \(d = 3\).

Step 3: Here, \(k = 5\).

$$\text{5th term from the end} = l - (5-1)d$$

$$= 26 - 4(3)$$

$$= 26 - 12$$

$$= 14$$

Answer: The 5th term from the end is 14.

Check by counting backward: 26, 23, 20, 17, 14.

Worked Example 3: When the AP is given by first term, common difference, and number of terms

An AP has first term \(a = 7\), common difference \(d = 5\), and 12 terms. Find the 4th term from the end.

Step 1: Find the last term.

Since there are 12 terms, the last term is:

$$l = a + (n-1)d$$

$$= 7 + (12-1)\times 5$$

$$= 7 + 11\times 5$$

$$= 7 + 55 = 62$$

Step 2: Use the formula for the 4th term from the end.

Here, \(k = 4\).

$$\text{4th term from the end} = l - (4-1)d$$

$$= 62 - 3(5)$$

$$= 62 - 15 = 47$$

Answer: The 4th term from the end is 47.

Worked Example 4: Using a decreasing AP

Find the 3rd term from the end of the AP:

$$30, 26, 22, 18, 14, 10$$

This is also an AP, but it decreases each time.

Step 1: Last term is \(l = 10\).

Step 2: Common difference is \(d = -4\).

Notice that the common difference must keep its sign. Since the sequence goes down by 4, \(d = -4\), not 4.

Step 3: Here, \(k = 3\).

$$\text{3rd term from the end} = l - (3-1)d$$

$$= 10 - 2(-4)$$

$$= 10 + 8$$

$$= 18$$

Answer: The 3rd term from the end is 18.

Check by counting backward: 10, 14, 18.

6. Common mistakes to avoid

• Using \(k\) instead of \(k-1\): For the kth term from the end, the formula is \(l - (k-1)d\), not \(l-kd\).
• Forgetting the sign of \(d\): In a decreasing AP, \(d\) is negative.
• Mixing up last term and first term: Terms from the end are based on the last term \(l\).
• Counting backward incorrectly: The 1st term from the end is the last term itself.

7. Quick check idea

After solving, try counting backward from the last term to see if your answer makes sense. This is a simple way to check your work.

8. Key formulas to remember

• nth term of an AP: $$T_n = a + (n-1)d$$
• Last term of a finite AP: $$l = a + (n-1)d$$
• kth term from the end: $$l - (k-1)d$$

Summary

To find a specific term from the end of a finite AP, start from the last term and move backward using the common difference.

If the last term is \(l\), then the kth term from the end is:

$$l - (k-1)d$$

This formula works for both increasing and decreasing APs. Just be careful to use the correct value of \(d\), including its sign.

Arithmetic Mean Between Numbers is about placing one or more numbers between two given numbers so that all the numbers form an arithmetic progression (AP).

In an arithmetic progression, the difference between consecutive terms is always the same. This constant difference is called the common difference.

For example, in the sequence \(2, 5, 8, 11\), the difference between each pair of neighboring terms is \(3\). So this is an arithmetic progression with common difference \(d = 3\).

When we talk about the arithmetic mean between numbers, we mean the number or numbers inserted between two given numbers so that the full list becomes an arithmetic progression.

If there is only one number between two numbers, that number is the arithmetic mean of the two numbers.

For example, between \(4\) and \(10\), the arithmetic mean is:

So \(4, 7, 10\) is an arithmetic progression.

But sometimes we need to insert more than one arithmetic mean. Then we must find the common difference first.

Main idea: If you insert \(n\) arithmetic means between two given numbers, then the total number of gaps is \(n+1\).

Suppose the first number is \(a\) and the last number is \(b\). Then:

Once you know \(d\), keep adding it to get each missing term.

Why does this formula work? If there are \(n\) numbers inserted, then the full sequence has:

• the first number,
• the \(n\) inserted numbers,
• and the last number.

That means there are \(n+2\) total terms, so there are \(n+1\) equal jumps from the first number to the last number.

The total change from first to last is \(b-a\). Since each jump is the same, we divide by the number of jumps:

Steps for inserting arithmetic means:

1. Write the first term \(a\) and the last term \(b\).
2. Count how many arithmetic means must be inserted. Call this number \(n\).
3. Find the common difference using \(d = \frac{b-a}{n+1}\).
4. Start at \(a\) and keep adding \(d\) to get the missing terms.
5. Check that the last term really becomes \(b\).

Special case: one arithmetic mean

If only one number is inserted between \(a\) and \(b\), then \(n=1\). So:

The middle number is:

So the arithmetic mean of two numbers is their average.

Worked Example 1: Find one arithmetic mean between 6 and 14

We need one number between \(6\) and \(14\).

Use the average:

So the arithmetic mean is \(10\).

The full arithmetic progression is:

Check the common difference:

The differences match, so the answer is correct.

Worked Example 2: Insert 3 arithmetic means between 8 and 24

We are inserting \(n=3\) means between \(8\) and \(24\).

Use the formula for the common difference:

Now start from \(8\) and add \(4\) each time:

• First term: \(8\)
• Next term: \(8+4=12\)
• Next term: \(12+4=16\)
• Next term: \(16+4=20\)
• Last term: \(20+4=24\)

So the 3 arithmetic means are:

The full AP is:

Worked Example 3: Insert 4 arithmetic means between \(-3\) and 12

This example includes a negative number, but the method stays the same.

Here, \(a=-3\), \(b=12\), and \(n=4\).

Find the common difference:

Now add \(3\) repeatedly starting from \(-3\):

So the 4 arithmetic means are:

Check:

All differences are equal, so the sequence is an AP.

Worked Example 4: Insert 2 arithmetic means between 5 and 2

Here the sequence is decreasing, so the common difference will be negative.

We have \(a=5\), \(b=2\), and \(n=2\).

Find the common difference:

Now add \(-1\) repeatedly:

So the 2 arithmetic means are:

This shows that arithmetic progressions can increase or decrease.

Using the nth term idea

Sometimes it helps to remember the nth term of an AP:

If you know the first and last terms and how many terms are in the full sequence, you can also use this formula to find \(d\).

For example, if 3 means are inserted between \(8\) and \(24\), then the full sequence has 5 terms:

Using \(a_1=8\):

This gives the same result as before.

Common mistakes to avoid

• Dividing by the wrong number: If you insert \(n\) means, divide by \(n+1\), not by \(n\).
• Forgetting the sign: If the sequence goes downward, the common difference is negative.
• Not checking the final term: After adding the common difference, make sure you reach the given last number exactly.
• Mixing up “terms” and “means”: Inserted means are only the missing middle numbers, not the full sequence.

Quick practice ideas

• Find one arithmetic mean between \(9\) and \(17\).
• Insert 2 arithmetic means between \(1\) and \(10\).
• Insert 3 arithmetic means between \(-8\) and \(4\).

Answers:

• Between \(9\) and \(17\): \(13\)
• Between \(1\) and \(10\): \(4, 7\)
• Between \(-8\) and \(4\): \(-5, -2, 1\)

Summary

An arithmetic mean is a number inserted between two numbers so that the numbers form an arithmetic progression.

If only one arithmetic mean is needed, use the average:

If \(n\) arithmetic means are inserted between \(a\) and \(b\), first find the common difference using:

Then keep adding \(d\) starting from the first number to find all the missing terms.

Deriving the Sum of the First n Terms of an Arithmetic Progression

An arithmetic progression (or AP) is a sequence of numbers where the same value is added each time to get the next term.

For example, in the sequence \(3, 7, 11, 15, 19\), each term increases by \(4\). So this is an arithmetic progression with:

• first term \(a = 3\)
• common difference \(d = 4\)

In this lesson, we will learn how to find the sum of the first \(n\) terms of an arithmetic progression. This sum is written as \(S_n\).

By the end, you will understand where the formulas come from and how to use these two important forms:

$$S_n = \frac{n}{2}\big(2a + (n-1)d\big)$$

and

$$S_n = \frac{n}{2}(a + l)$$

where:

• \(a\) = first term
• \(d\) = common difference
• \(n\) = number of terms
• \(l\) = last term

1. What does the sum of the first \(n\) terms mean?

If an arithmetic progression is \(2, 5, 8, 11, 14\), then the sum of the first 5 terms is:

$$S_5 = 2 + 5 + 8 + 11 + 14 = 40$$

When there are many terms, adding one by one takes too long. So we use a formula.

2. Writing the terms of an arithmetic progression

If the first term is \(a\) and the common difference is \(d\), then the terms are:

$$a,\ a+d,\ a+2d,\ a+3d,\ \dots$$

The \(n\)th term is:

$$T_n = a + (n-1)d$$

If we call the last term \(l\), then:

$$l = a + (n-1)d$$

This result is very important because it helps us connect the two sum formulas.

3. Deriving the formula for \(S_n\)

Suppose we want to add the first \(n\) terms of an arithmetic progression:

$$S_n = a + (a+d) + (a+2d) + \dots + \big(a+(n-1)d\big)$$

Now write the same sum again, but in reverse order:

$$S_n = \big(a+(n-1)d\big) + \big(a+(n-2)d\big) + \dots + (a+d) + a$$

Now add the two lines term by term.

The first terms add to:

$$a + \big(a+(n-1)d\big) = 2a + (n-1)d$$

The second terms add to:

$$ (a+d) + \big(a+(n-2)d\big) = 2a + (n-1)d$$

Every pair has the same sum. Since there are \(n\) pairs, we get:

$$2S_n = n\big(2a + (n-1)d\big)$$

Now divide both sides by 2:

$$S_n = \frac{n}{2}\big(2a + (n-1)d\big)$$

This is the first formula for the sum of the first \(n\) terms of an arithmetic progression.

4. Deriving the second form using the last term

We know that the last term is:

$$l = a + (n-1)d$$

Substitute this into the first sum formula:

$$S_n = \frac{n}{2}\big(a + a + (n-1)d\big)$$

$$S_n = \frac{n}{2}(a + l)$$

So the second formula is:

$$S_n = \frac{n}{2}(a + l)$$

This means:

Sum = number of terms × average of first and last term

That makes sense because in an arithmetic progression, the terms are evenly spaced, so the average of the first and last term is the average of all the terms.

5. When should you use each formula?

• Use $$S_n = \frac{n}{2}\big(2a + (n-1)d\big)$$ when you know \(a\), \(d\), and \(n\).
• Use $$S_n = \frac{n}{2}(a + l)$$ when you know \(a\), \(l\), and \(n\).

If you are not given the last term, you can find it first using:

$$l = a + (n-1)d$$

Worked Example 1: Find the sum of the first 10 terms

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

$$4, 7, 10, 13, \dots$$

Step 1: Identify the values

• First term: \(a = 4\)
• Common difference: \(d = 3\)
• Number of terms: \(n = 10\)

Step 2: Use the formula

$$S_n = \frac{n}{2}\big(2a + (n-1)d\big)$$

Substitute the values:

$$S_{10} = \frac{10}{2}\big(2(4) + (10-1)(3)\big)$$

$$S_{10} = 5(8 + 27)$$

$$S_{10} = 5(35) = 175$$

Answer: The sum of the first 10 terms is \(175\).

Worked Example 2: Use the first and last term

Find the sum of the arithmetic progression:

$$6, 10, 14, 18, 22, 26$$

Step 1: Identify the values

• First term: \(a = 6\)
• Last term: \(l = 26\)
• Number of terms: \(n = 6\)

Step 2: Use the formula

$$S_n = \frac{n}{2}(a + l)$$

Substitute:

$$S_6 = \frac{6}{2}(6 + 26)$$

$$S_6 = 3(32) = 96$$

Answer: The sum is \(96\).

Worked Example 3: Find the last term first

Find the sum of the first 15 terms of the arithmetic progression with:

• first term \(a = 2\)
• common difference \(d = 5\)

Step 1: Find the 15th term

$$l = a + (n-1)d$$

$$l = 2 + (15-1)(5)$$

$$l = 2 + 14 \cdot 5$$

$$l = 2 + 70 = 72$$

Step 2: Use the sum formula

$$S_n = \frac{n}{2}(a + l)$$

$$S_{15} = \frac{15}{2}(2 + 72)$$

$$S_{15} = \frac{15}{2}(74)$$

$$S_{15} = 15 \cdot 37 = 555$$

Answer: The sum of the first 15 terms is \(555\).

Worked Example 4: Find the number of terms from a sum

The arithmetic progression starts:

$$3, 8, 13, 18, \dots$$

The sum of the first \(n\) terms is \(273\). Find \(n\).

Step 1: Identify the values

• \(a = 3\)
• \(d = 5\)
• \(S_n = 273\)

Step 2: Use the formula

$$S_n = \frac{n}{2}\big(2a + (n-1)d\big)$$

Substitute:

$$273 = \frac{n}{2}\big(2(3) + (n-1)(5)\big)$$

$$273 = \frac{n}{2}(6 + 5n - 5)$$

$$273 = \frac{n}{2}(5n + 1)$$

Multiply both sides by 2:

$$546 = n(5n + 1)$$

$$5n^2 + n - 546 = 0$$

Now solve the equation. We look for two numbers that multiply to \(5 \times -546 = -2730\) and add to \(1\). Those numbers are \(66\) and \(-65\).

$$5n^2 + 66n - 65n - 546 = 0$$

$$n(5n + 66) - 13(5n + 66) = 0$$

$$ (n - 13)(5n + 66) = 0$$

So:

$$n = 13 \quad \text{or} \quad 5n + 66 = 0$$

$$n = 13 \quad \text{or} \quad n = -\frac{66}{5}$$

A number of terms cannot be negative, so:

Answer: \(n = 13\)

6. Common mistakes to avoid

• Mixing up \(n\) and \(d\): \(n\) is the number of terms, and \(d\) is the common difference.
• Using the wrong last term: Remember, the last term is \(l = a + (n-1)d\), not \(a + nd\).
• Forgetting brackets: In formulas like \(\frac{n}{2}(a+l)\), add inside the brackets first.
• Using the formula on a non-arithmetic sequence: The difference between terms must be constant.

7. Quick check for understanding

1. What is the common difference in \(9, 13, 17, 21, \dots\)?
2. Find the sum of the first 8 terms of \(5, 8, 11, 14, \dots\).
3. If \(a = 12\), \(d = 6\), and \(n = 7\), find the last term.
4. Then find the sum of those 7 terms.

Answers:

1. \(d = 4\)
2. $$S_8 = \frac{8}{2}\big(2(5) + (8-1)(3)\big) = 4(10+21)=124$$
3. $$l = 12 + (7-1)6 = 12 + 36 = 48$$
4. $$S_7 = \frac{7}{2}(12+48)=\frac{7}{2}(60)=210$$

Summary

To find the sum of the first \(n\) terms of an arithmetic progression, we use:

$$S_n = \frac{n}{2}\big(2a + (n-1)d\big)$$

or

$$S_n = \frac{n}{2}(a + l)$$

The first formula is useful when you know the first term, common difference, and number of terms. The second formula is useful when you know the first term, last term, and number of terms.

The key idea in the derivation is that if you write the sum forward and backward, each pair adds to the same value. That makes it possible to build a quick formula instead of adding every term one at a time.

Relationship Between the nth Term and the Sum of n Terms

In an arithmetic progression (AP), numbers increase or decrease by the same amount each time. For example, in the sequence 3, 7, 11, 15, ... the common difference is 4.

Sometimes a question gives us the sum of the first n terms, written as \(S_n\), instead of the sequence itself. In that case, we may still need to find the nth term, written as \(a_n\).

The key idea is very important:

When we subtract the sum of the first \((n-1)\) terms from the sum of the first \(n\) terms, everything cancels except the nth term.

So the relationship is:

This formula works because:

and

Now subtract:

This is the main relationship you need to remember for this lesson.

Why this matters in arithmetic progressions

In AP questions, you may be given a formula for \(S_n\), such as \(S_n = 3n^2 + 2n\), and asked to find the nth term. Since \(S_n\) tells us the total up to term \(n\), we use subtraction to isolate just one term.

This method is especially useful when:

• the sum formula is given directly,
• you need to find a particular term,
• or you want to check whether a sum formula produces an arithmetic progression.

Step-by-step method

1. Write down the given formula for \(S_n\).
2. Find \(S_{n-1}\) by replacing every \(n\) with \((n-1)\).
3. Use the formula \(a_n = S_n - S_{n-1}\).
4. Simplify carefully.

Be very careful when substituting \((n-1)\), especially if there are brackets or powers.

Worked Example 1: Finding the nth term from a simple sum formula

Suppose

Find \(a_n\).

Step 1: Write \(S_n\).

Step 2: Find \(S_{n-1}\).

Now expand:

Step 3: Subtract.

Answer: The nth term is

Check: Find the first few terms.

Using \(a_n = 4n + 1\):

• \(a_1 = 5\)
• \(a_2 = 9\)
• \(a_3 = 13\)

This is an arithmetic progression with common difference 4.

Worked Example 2: Finding a particular term

Suppose

Find the 10th term.

First find the general nth term.

Find \(S_{n-1}\):

Now subtract:

Now find the 10th term:

Answer: The 10th term is \(24\).

Worked Example 3: Showing the sequence is an AP

Suppose

Find \(a_n\) and show that the sequence is an arithmetic progression.

First find \(S_{n-1}\):

Now subtract:

Now find the first few terms:

• \(a_1 = 8(1) - 5 = 3\)
• \(a_2 = 8(2) - 5 = 11\)
• \(a_3 = 8(3) - 5 = 19\)

The differences are:

• \(11 - 3 = 8\)
• \(19 - 11 = 8\)

Since the difference is constant, the sequence is an arithmetic progression.

Worked Example 4: Starting from values of sums

A student is told that:

Find the 5th term.

Use the relationship directly:

Answer: The 5th term is \(14\).

This shows that the formula is not only for algebra expressions like \(S_n\), but also for actual sum values.

Important pattern to notice

If the sum formula \(S_n\) is a quadratic expression in \(n\), then the nth term \(a_n\) usually becomes a linear expression in \(n\). A linear nth term means the sequence has a constant difference, which matches an arithmetic progression.

For example:

• If \(S_n = 2n^2 + 3n\), then \(a_n = 4n + 1\).
• If \(S_n = n^2 + 5n\), then \(a_n = 2n + 4\).

This is a useful clue in many exam questions.

Common mistakes to avoid

• Using \(S_n\) as the nth term: \(S_n\) is the sum of the first \(n\) terms, not the nth term itself.
• Forgetting brackets: When finding \(S_{n-1}\), always write brackets, like \((n-1)^2\).
• Subtracting incorrectly: In \(S_n - S_{n-1}\), every term inside the second bracket changes sign.
• Mixing up \(a_n\) and \(a_{n-1}\): The formula gives the nth term exactly: \(a_n = S_n - S_{n-1}\).

Quick practice ideas

If you want to practise, try these on your own:

• If \(S_n = 3n^2 + n\), find \(a_n\).
• If \(S_n = 5n^2 - 2n\), find the 6th term.
• If \(S_8 = 120\) and \(S_7 = 99\), find \(a_8\).

Brief Summary

The most important relationship is:

This works because the sum of the first \((n-1)\) terms is contained inside the sum of the first \(n\) terms, so subtracting leaves only the nth term.

To use it, first find \(S_{n-1}\), then subtract from \(S_n\), and simplify carefully. This is a powerful method for finding terms of an arithmetic progression when the sum formula is given.

Properties of Arithmetic Progressions

An arithmetic progression, or AP, is a sequence of numbers in which the difference between consecutive terms is always the same.

For example, the sequence \(3, 7, 11, 15, 19\) is an AP because each term increases by \(4\). The constant difference is called the common difference.

If the first term is \(a\) and the common difference is \(d\), then the AP looks like:

The \(n\)th term of an AP is:

In this lesson, we will study an important set of properties:

• What happens if we add the same number to every term of an AP?
• What happens if we subtract the same number from every term?
• What happens if we multiply every term by the same number?
• What happens if we divide every term by the same non-zero number?

We will show that in each case, the new sequence is still an arithmetic progression.

Why this matters: These properties help us recognize APs in different forms and make it easier to solve sequence problems.

1. Adding a constant to every term of an AP

Suppose we start with an AP:

Now add a constant number \(k\) to every term. The new sequence becomes:

To check whether this is an AP, look at the difference between consecutive terms:

The difference is still constant, so the new sequence is also an AP.

Conclusion: Adding the same constant to every term of an AP gives another AP with the same common difference.

2. Subtracting a constant from every term of an AP

Now subtract a constant \(k\) from every term:

Again, find the difference between consecutive terms:

The common difference is still \(d\), so this new sequence is also an AP.

Conclusion: Subtracting the same constant from every term of an AP gives another AP with the same common difference.

3. Multiplying every term of an AP by a constant

Now multiply every term by a constant \(k\):

Simplifying, we get:

Now check the differences:

The difference is constant, so this is also an AP.

Conclusion: Multiplying every term of an AP by a constant gives another AP. The new common difference becomes \(kd\).

If \(k\) is negative, the sequence may decrease instead of increase, but it is still an AP because the difference remains constant.

4. Dividing every term of an AP by a non-zero constant

Now divide every term by a non-zero constant \(k\):

This can be written as:

Now the difference between consecutive terms is:

This is constant, so the new sequence is also an AP.

Conclusion: Dividing every term of an AP by the same non-zero constant gives another AP. The new common difference becomes \(\frac{d}{k}\).

We say non-zero because division by zero is not defined.

General idea behind all four properties

An AP stays an AP when every term is changed in the same way by a constant operation. The key reason is that the difference between one term and the next stays constant.

Let the original AP have terms \(T_1, T_2, T_3,\dots\) with common difference \(d\). Then:

If we create a new sequence by adding \(k\), subtracting \(k\), multiplying by \(k\), or dividing by \(k\neq 0\), the differences in the new sequence are still constant.

This is the main property we use to prove the new sequence is also an AP.

Worked Example 1: Adding a constant

Consider the AP:

The common difference is:

Now add \(4\) to every term:

Check the new differences:

The new sequence is still an AP, and the common difference remains \(3\).

Worked Example 2: Subtracting and multiplying

Start with the AP:

This is an AP with common difference:

Part A: Subtract 2 from every term

Check the difference:

So the new sequence is still an AP with common difference \(-4\).

Part B: Multiply every term by 3

Now check the difference:

This is also an AP. The new common difference is:

Worked Example 3: Dividing by a constant

Take the AP:

The common difference is:

Divide every term by \(3\):

Check the new difference:

The new sequence is an AP with common difference:

Worked Example 4: Using the formula

The \(n\)th term of an AP is:

So the first term is \(a=7\) and the common difference is \(d=5\).

Now multiply every term by \(-2\). The new \(n\)th term becomes:

Simplify:

This is in the form of an arithmetic progression, with:

• first term \(-14\)
• common difference \(-10\)

So multiplying all terms by \(-2\) gives a new AP.

Important observations

• Adding or subtracting a constant changes the first term, but not the common difference.
• Multiplying or dividing by a constant changes both the first term and the common difference.
• If you multiply by a negative number, the AP may switch from increasing to decreasing, or from decreasing to increasing.
• Dividing by zero is not allowed.

Quick comparison table

• Original AP: first term \(a\), common difference \(d\)
• Add \(k\): new first term \(a+k\), new difference \(d\)
• Subtract \(k\): new first term \(a-k\), new difference \(d\)
• Multiply by \(k\): new first term \(ka\), new difference \(kd\)
• Divide by \(k\neq 0\): new first term \(\frac{a}{k}\), new difference \(\frac{d}{k}\)

How to prove a new sequence is an AP

When you are asked to prove a sequence is an AP, use this method:

1. Write the original AP in general form: \(a, a+d, a+2d,\dots\)
2. Apply the operation to every term.
3. Find the difference between two consecutive terms in the new sequence.
4. If the difference is constant, the sequence is an AP.

Practice thought

If you start with the AP \(2, 5, 8, 11, \dots\):

• adding \(10\) gives \(12, 15, 18, 21,\dots\)
• subtracting \(1\) gives \(1, 4, 7, 10,\dots\)
• multiplying by \(4\) gives \(8, 20, 32, 44,\dots\)
• dividing by \(2\) gives \(1, 2.5, 4, 5.5,\dots\)

Each new sequence still has a constant difference, so each is an AP.

Summary

An arithmetic progression is a sequence with a constant difference between consecutive terms. If we add or subtract the same constant to every term, the result is still an AP with the same common difference.

If we multiply every term by a constant, the result is also an AP, but the common difference is multiplied by that constant. If we divide every term by a non-zero constant, the result is again an AP, and the common difference is divided by that constant.

The reason all of these work is simple: the difference between one term and the next remains constant.

Modeling Discrete Linear Growth Patterns means using arithmetic progressions to describe situations where a quantity changes by the same amount each step.

These patterns are called discrete because they happen in separate steps, such as each week, each month, or each row. They are called linear growth patterns because the change is constant.

In real life, this can happen when:

• you save a fixed amount more each week than the week before,
• loan payments increase by a fixed amount each month,
• objects are stacked so that each new layer has a fixed number more than the previous layer.

To model these situations, we use the formulas for an arithmetic progression (AP).

An arithmetic progression is a sequence where the difference between consecutive terms is constant.

If the first term is \(a_1\) and the common difference is \(d\), then the sequence looks like:

$$a_1,\ a_1+d,\ a_1+2d,\ a_1+3d,\dots$$

The formula for the \(n\)th term is:

$$a_n=a_1+(n-1)d$$

This formula helps us find the value at any step without listing all the terms.

Sometimes, we also need the total of several terms. For example, we may want the total amount saved after 10 weeks or the total number of blocks in several rows.

The sum of the first \(n\) terms is:

$$S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)$$

Another useful form is:

$$S_n=\frac{n}{2}(a_1+a_n)$$

We use this when we know the first and last terms.

Main idea: when a situation increases or decreases by the same amount each step, we can model it with an arithmetic progression.

To model a real-world problem, follow these steps:

1. Identify what changes each step.
2. Find the first term \(a_1\).
3. Find the common difference \(d\).
4. Use \(a_n=a_1+(n-1)d\) to find a specific term.
5. Use a sum formula if you need the total across many steps.

Let us now look at some common contexts.

1. Sequential savings

Suppose a student saves money every week. In week 1, they save R50. Each week, they save R10 more than the previous week.

The amounts saved each week form this sequence:

$$50,\ 60,\ 70,\ 80,\dots$$

Here:

• \(a_1=50\)
• \(d=10\)

The amount saved in week \(n\) is:

$$a_n=50+(n-1)(10)$$

If we want the amount saved in week 8:

$$a_8=50+(8-1)(10)=50+70=120$$

So, the student saves R120 in week 8.

If we want the total saved in 8 weeks, we use the sum formula:

$$S_8=\frac{8}{2}\left(2(50)+(8-1)(10)\right)$$

$$S_8=4(100+70)=4(170)=680$$

So, the total saved after 8 weeks is R680.

2. Loan repayments that change by a fixed amount

Sometimes a repayment plan is designed so that the payment increases by the same amount each month.

For example, suppose a person pays R300 in the first month and then increases the payment by R25 each month.

The payments form an arithmetic progression:

$$300,\ 325,\ 350,\ 375,\dots$$

Here:

• \(a_1=300\)
• \(d=25\)

The payment in month \(n\) is:

$$a_n=300+(n-1)(25)$$

If we want the payment in month 12:

$$a_{12}=300+(12-1)(25)=300+275=575$$

So, the payment in month 12 is R575.

If we want the total paid over 12 months:

$$S_{12}=\frac{12}{2}(300+575)$$

$$S_{12}=6(875)=5250$$

So, the total amount paid over 12 months is R5250.

3. Geometric stacking

Arithmetic progressions can also model physical arrangements, such as stacks of cans, blocks, or seats arranged in rows.

For example, imagine a display made of rows of boxes. The first row has 4 boxes. Each new row has 3 more boxes than the row before.

The numbers of boxes in each row are:

$$4,\ 7,\ 10,\ 13,\dots$$

This is an arithmetic progression with:

• \(a_1=4\)
• \(d=3\)

The number of boxes in row \(n\) is:

$$a_n=4+(n-1)(3)$$

If there are 9 rows, then the 9th row has:

$$a_9=4+(9-1)(3)=4+24=28$$

So, the 9th row has 28 boxes.

To find the total number of boxes in 9 rows:

$$S_9=\frac{9}{2}(4+28)$$

$$S_9=\frac{9}{2}(32)=144$$

So, the full display contains 144 boxes.

Worked Example 1: Finding a term from a real-life pattern

A learner saves R20 in the first week and increases the amount by R5 each week. How much do they save in week 15?

Step 1: Identify the first term and common difference.

• \(a_1=20\)
• \(d=5\)

Step 2: Use the nth-term formula.

$$a_n=a_1+(n-1)d$$

$$a_{15}=20+(15-1)(5)$$

$$a_{15}=20+14\cdot 5$$

$$a_{15}=20+70=90$$

Answer: The learner saves R90 in week 15.

Worked Example 2: Finding a total from a savings plan

A person saves R100 in the first month and then increases the amount by R20 each month. How much do they save in total over 6 months?

Step 1: Identify the sequence information.

• \(a_1=100\)
• \(d=20\)
• \(n=6\)

Step 2: Find the 6th term.

$$a_6=100+(6-1)(20)=100+100=200$$

Step 3: Use the sum formula.

$$S_6=\frac{6}{2}(100+200)$$

$$S_6=3(300)=900$$

Answer: The total saved over 6 months is R900.

Worked Example 3: Modeling a stacking pattern

A pyramid of cans has 6 cans in the first row, 8 cans in the second row, 10 cans in the third row, and so on. How many cans are in the 12th row, and how many cans are there in total in the first 12 rows?

Step 1: Identify the arithmetic progression.

The sequence is:

$$6,\ 8,\ 10,\ 12,\dots$$

• \(a_1=6\)
• \(d=2\)

Step 2: Find the 12th term.

$$a_{12}=6+(12-1)(2)=6+22=28$$

So, the 12th row has 28 cans.

Step 3: Find the total in the first 12 rows.

$$S_{12}=\frac{12}{2}(6+28)$$

$$S_{12}=6(34)=204$$

Answer: The 12th row has 28 cans, and the first 12 rows contain 204 cans in total.

Worked Example 4: Loan repayment model

A borrower pays R450 in the first month of a repayment plan. Each month, the payment increases by R30. Find:

1. the payment in month 10,
2. the total paid in the first 10 months.

Step 1: Identify the values.

• \(a_1=450\)
• \(d=30\)
• \(n=10\)

Step 2: Find the 10th payment.

$$a_{10}=450+(10-1)(30)$$

$$a_{10}=450+270=720$$

So, the payment in month 10 is R720.

Step 3: Find the total paid.

$$S_{10}=\frac{10}{2}(450+720)$$

$$S_{10}=5(1170)=5850$$

Answer:

• Month 10 payment: R720
• Total paid in 10 months: R5850

How to tell if a situation is an arithmetic progression

Ask yourself: does the amount increase or decrease by the same number each time?

If the answer is yes, then it can be modeled by an arithmetic progression.

For example:

• add 5 each week -> arithmetic progression,
• subtract 2 each month -> arithmetic progression,
• double each time -> not an arithmetic progression.

Common mistakes to avoid

• Mixing up \(n\) and \(n-1\): In the formula \(a_n=a_1+(n-1)d\), always use \(n-1\), not just \(n\).
• Using the wrong first term: Make sure \(a_1\) is the value at step 1, not step 0 or step 2.
• Forgetting what the question asks: Sometimes you need one term, and sometimes you need the total.
• Wrong common difference: Check the amount added or subtracted between terms carefully.

Quick check strategy

1. Write the first few terms if possible.
2. Check the constant difference.
3. Decide whether the question asks for a single term or a sum.
4. Choose the correct formula.
5. Substitute carefully and simplify step by step.

Summary

Discrete linear growth patterns happen when a quantity changes by a constant amount at each step. These situations can be modeled using arithmetic progressions.

Use:

$$a_n=a_1+(n-1)d$$

to find a specific term, and use:

$$S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)$$

or

$$S_n=\frac{n}{2}(a_1+a_n)$$

to find a total.

These formulas are useful in real-world contexts like savings plans, loan repayments, and stacking problems. The key is to recognize the constant change and match it to an arithmetic progression model.