Lesson Explainer: Arithmetic Series | Nagwa Lesson Explainer: Arithmetic Series | Nagwa

Lesson Explainer: Arithmetic Series Mathematics • Second Year of Secondary School

In this explainer, we will learn how to calculate the sum of the terms in an arithmetic sequence with a definite number of terms.

Sequences and series are commonly found in nature and can be used to model the spread of a virus or the decline of a population (these two things are not necessarily linked of course!). In the study of pure mathematics, we interest ourselves in finding the general, or 𝑛th, term of such sequences, as well as the sum of a given number of terms.

We begin by recalling what we mean by an arithmetic sequence.

Definition: Arithmetic Sequences and Series

An arithmetic sequence is a sequence that has a common difference between consecutive terms. The general term, π‘‡οŠ, of an arithmetic sequence with first term 𝑇 and common difference 𝑑 is given by 𝑇=𝑇+(π‘›βˆ’1)𝑑.

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

For instance, the sequence βˆ’4,βˆ’1,2,5,8,… is an example of an arithmetic sequence.

It has a first term βˆ’4 and a common difference of 3.

By using the general term with 𝑇=βˆ’4 and 𝑑=3, the 𝑛th term of this sequence is given by 𝑇=βˆ’4+(π‘›βˆ’1)Γ—3=βˆ’4+3π‘›βˆ’3=3π‘›βˆ’7.

The corresponding arithmetic series would be βˆ’4+βˆ’1+2+5+8+β‹―.

Let’s look at a practical example before deriving a formula for the sum of a given number of terms in an arithmetic series.

Example 1: Finding the Sum of an Arithmetic Series given the First Three Terms

Find the sum of the first 17 terms of the arithmetic series 12+21+30+β‹―.

Answer

This is an arithmetic series with first term 12. The common difference, 𝑑, is found by subtracting a term by the term that precedes it: 𝑑=21βˆ’12=9.

The general term, π‘‡οŠ, of an arithmetic sequence with first term 𝑇 and common difference 𝑑 is given by 𝑇=𝑇+(𝑛+1)𝑑.

This means the general term of our series is 𝑇=12+(π‘›βˆ’1)Γ—9=12+9π‘›βˆ’9=9𝑛+3.

The final two terms in the partial sum of this series are found by letting 𝑛=16 and 𝑛=17.

When 𝑛=16, 𝑇=9Γ—16+3=147.

When 𝑛=17, 𝑇=9Γ—17+3=156.

This means the sum of the first 17 terms, π‘†οŠ§οŠ­, can be written as 𝑆=12+21+30+β‹―+147+156.

Of course, we could reverse the series and still get the same result: 𝑆=156+147+β‹―+30+21+12.

Notice how each of the 17 numbers in the list can be paired up with a number in the other list to give a sum of 168.

Adding these equations gives 2𝑆=168Γ—17=2856.

The sum of the first 17 terms can now be found by dividing 2β€Žβ€‰β€Ž856 by 2: 𝑆=28562=1428.

The sum of the first 17 terms of this arithmetic series is 1β€Žβ€‰β€Ž428.

This method for finding the sum of a finite arithmetic series can be generalized to an arithmetic series with first term 𝑇 and common difference 𝑑.

Example 2: Writing an Expression for the Sum of an Arithmetic Sequence

Find an expression for the sum of an arithmetic sequence whose first term is 𝑇 and whose common difference is 𝑑.

Answer

We recall that the general term, π‘‡οŠ, of an arithmetic sequence with first term 𝑇 and common difference 𝑑is given by 𝑇=𝑇+(π‘›βˆ’1)𝑑.

We can use this formula to work out the first 𝑛 terms in this sequence.

When 𝑛=1, 𝑇=𝑇+(1βˆ’1)𝑑=𝑇.

When 𝑛=2, 𝑇=𝑇+(2βˆ’1)𝑑=𝑇+𝑑.

When 𝑛=3, 𝑇=𝑇+(3–1)𝑑=𝑇+2𝑑.

The pattern continues in this manner.

The sum of the first 𝑛 terms, π‘†οŠ, is now given as 𝑆=𝑇+𝑇+𝑑+𝑇+2𝑑+β‹―+𝑇+(𝑛–2)𝑑+𝑇+(𝑛–1)𝑑.

Of course, if we reverse the series, we will still get the same overall sum: 𝑆=𝑇+(𝑛–1)𝑑+𝑇+(𝑛–2)𝑑+β‹―+𝑇+2𝑑+𝑇+𝑑+𝑇.

Notice how each number in the list can be paired up with a number in the other list to give a constant sum:

𝑇+𝑇+(𝑛–1)𝑑=2𝑇+(𝑛–1)𝑑.

This means that when we add the two equations, we will have 𝑛 lots of this expression: 2𝑆=𝑛×(2𝑇+(𝑛–1)𝑑).

To find an expression for π‘†οŠ, we then divide through by 2: 𝑆=𝑛2(2𝑇+(𝑛–1)𝑑).

Definition: The Sum of an Arithmetic Sequence

The sum of the first 𝑛 terms of an arithmetic series whose first term is 𝑇 and whose common difference is 𝑑 is given by π‘†οŠ, where 𝑆=𝑛2(2𝑇+(𝑛–1)𝑑).

There will be occasions when we are given the first and last terms of an arithmetic series and asked to calculate its sum. We will now consider how to derive a formula for this sum using the formula for the sum of the first 𝑛 terms.

Example 3: Writing an Expression for the Sum of the First 𝑛 Terms of an Arithmetic Sequence

Write an expression for the sum of the first 𝑛 terms of an arithmetic sequence with first term 𝑇 and last term 𝑙.

Answer

We begin by recalling the formula that allows us to find the 𝑛th term of an arithmetic series with first term 𝑇 and common difference 𝑑: 𝑇=𝑇+(𝑛–1)𝑑.

We also know that the sum of the first 𝑛th terms of an arithmetic series whose first term is 𝑇 and whose common difference is 𝑑 is given by π‘†οŠ, where 𝑆=𝑛2(2𝑇+(𝑛–1)𝑑).

By writing 2𝑇 as 𝑇+𝑇, we can substitute π‘‡οŠ in, as shown: 𝑆=𝑛2(𝑇+𝑇+(𝑛–1)𝑑)=𝑛2(𝑇+𝑇).

In a sequence with 𝑛 terms, π‘‡οŠ is the last term. This means we can replace π‘‡οŠ with 𝑙 to find a formula for the sum of the first 𝑛 terms of an arithmetic sequence with first term 𝑇 and last term 𝑙: 𝑆=𝑛2(𝑇+𝑙).

Definition: The Sum of an Arithmetic Sequence

The sum of the first 𝑛 terms of an arithmetic sequence with first term 𝑇 and last term 𝑙 is given by π‘†οŠ, where 𝑆=𝑛2(𝑇+𝑙).

We will now learn how to apply this formula to find the sum of the terms in a finite arithmetic series.

Example 4: Finding the Sum of an Arithmetic Series given the First and Last Terms

Find the sum of the terms of the 11-term arithmetic sequence whose first term is βˆ’92 and last term is βˆ’102.

Answer

Recall that the sum of the first 𝑛 terms of an arithmetic sequence with first term 𝑇 and last term 𝑙 is given by π‘†οŠ, where 𝑆=𝑛2(𝑇+𝑙).

The first term in our sequence is βˆ’92 and the last term is βˆ’102, so we will let 𝑇=βˆ’92 and 𝑙=βˆ’102.

There are 11 terms in the sequence, so we will let 𝑛=11.

Then, the sum of the first 11 terms is given by π‘†οŠ§οŠ§, where 𝑆=112(βˆ’92+(βˆ’102))=112(βˆ’194)=βˆ’1067.

The sum of the terms of this 11-term arithmetic sequence is βˆ’1067.

We will now investigate how we can use the general term of an arithmetic sequence to calculate the sum of a given number of terms of the same arithmetic sequence.

Example 5: Finding the Sum of an Arithmetic Sequence given the General Term

Find the sum of the first 10 terms of the sequence π‘‡οŠ, where 𝑇=2𝑛+4.

Answer

We are given the general term of the sequence, 𝑇=2𝑛+4. This is a formula that allows us to calculate any term given its position number.

For instance, the first term is found by substituting 𝑛=1 into the formula.

When 𝑛=1, 𝑇=2Γ—1+4=6.

When 𝑛=2, 𝑇=2Γ—2+4=8.

When 𝑛=3, 𝑇=2Γ—3+4=10.

The first three terms of the sequence are 6, 8, and 10. We can therefore deduce the first term to be 6 and the common difference to be 2.

The sum of the first 𝑛 terms of an arithmetic series whose first term is 𝑇 and whose common difference is 𝑑 is given by π‘†οŠ, where 𝑆=𝑛2(2𝑇+(𝑛–1)𝑑).

Since we are calculating the sum of the first 10 terms, we will let 𝑛=10, 𝑇=6, and 𝑑=2: 𝑆=102(2Γ—6+(10–1)Γ—2)=5(12+18)=150.

The sum of the first 10 terms of this sequence is 150.

It is worth noting that the formulae for working with sequences and series can be adapted when a term or a number of terms are given as algebraic expressions. In our next example, we will see what that could look like.

Example 6: Finding the Sum of a Given Arithmetic Sequence in terms of 𝑛

Find, in terms of 𝑛, the sum of the arithmetic sequence 9,10,11,…,𝑛+8.

Answer

In order to find the sum of an arithmetic sequence, we need to know either the number of terms or the value of the last term. In this example, we are given the last term of the arithmetic sequence as an algebraic expression, 𝑛+8.

In order to establish the number of the terms in this sequence, let’s look at the general term 𝑛+8. When 𝑛=1, 𝑛+8=9, which is the first term in the sequence. When 𝑛=2, 𝑛+8=10, which is the second term in the sequence. This pattern continues, meaning that, as long as 𝑛β‰₯1, 𝑛 is also the number of terms in the sequence.

This means we can use the formula to find the sum of the first 𝑛 terms of an arithmetic sequence with first term 𝑇 and last term 𝑙: 𝑆=𝑛2(𝑇+𝑙).

Substituting 𝑇=9 and 𝑙=𝑛+8 into our formula, 𝑆=𝑛2(9+𝑛+8)=𝑛2(𝑛+17).

The sum of the arithmetic sequence 9,10,11,…,𝑛+8 is 𝑛2(𝑛+17).

In our final example, we look at how we find the sum of a given number of terms of an arithmetic sequence given information about its terms. This process will involve some problem solving to β€œwork backward” to a solution.

Example 7: Finding the Sum of a Given Number of Terms of an Arithmetic Sequence under a Given Condition

Find the sum of the first 21 terms of an arithmetic sequence given 𝑇+𝑇=βˆ’232οŠͺ and 𝑇=βˆ’130.

Answer

There are two formulae that we can use to find the sum of a given number of terms of an arithmetic sequence. The first requires us to know the value of the first term, 𝑇, and the common difference, 𝑑: 𝑆=𝑛2(2𝑇+(π‘›βˆ’1)𝑑), while the second requires us to know the value of the first term, 𝑇, and the last term, 𝑙: 𝑆=𝑛2(𝑇+𝑙).

We have been given information about three of the terms in the sequence, so it follows that we might need to apply the formula for the 𝑛th term of an arithmetic sequence. Since this formula uses the value of the first term and the common difference, we might deduce that we will need the first version of the summation formula.

The general term, π‘‡οŠ, of an arithmetic sequence with first term 𝑇 and common difference 𝑑 is given by 𝑇=𝑇+(𝑛–1)𝑑.

Letting 𝑛=41, 𝑇=𝑇+(41–1)𝑑=𝑇+40𝑑.οŠͺ

For 𝑛=9, 𝑇=𝑇+(9–1)𝑑=𝑇+8𝑑.

And for 𝑛=27, 𝑇=𝑇+(27–1)𝑑=𝑇+26𝑑.

Let’s substitute each of these expressions into the two equations given to us in the question.

The first gives us 𝑇+𝑇=βˆ’232𝑇+40𝑑+𝑇+8𝑑=βˆ’2322𝑇+48𝑑=βˆ’232𝑇+24𝑑=βˆ’116.οŠͺ

Then, the equation 𝑇=βˆ’130 becomes 𝑇+26𝑑=βˆ’130.

Notice that we have a pair of linear simultaneous equations. These can be solved by subtracting one from the other: 𝑇+26𝑑=βˆ’130βˆ’π‘‡+24𝑑=βˆ’1162𝑑=βˆ’14𝑑=βˆ’7.

Finally, we can substitute 𝑑=βˆ’7 into either of our original equations: 𝑇+24Γ—(βˆ’7)=βˆ’116𝑇–168=βˆ’116𝑇=52.

The sum of the first 𝑛 terms of an arithmetic series whose first term is 𝑇 and whose common difference is 𝑑 is given by π‘†οŠ, where 𝑆=𝑛2(2𝑇+(𝑛–1)𝑑).

Since we are finding the sum of the first 21 terms, we will substitute 𝑛=21, 𝑇=52, and 𝑑=βˆ’7 into this formula: 𝑆=212(2Γ—52+(21–1)Γ—(βˆ’7))=212(104–140)=βˆ’378.

Key Points

  • The sum of the first 𝑛 terms of an arithmetic series whose first term is 𝑇 and whose common difference is 𝑑 is given by π‘†οŠ, where 𝑆=𝑛2(2𝑇+(π‘›βˆ’1)𝑑).
  • The sum of the first 𝑛 terms of an arithmetic sequence with first term 𝑇 and last term 𝑙 is given by π‘†οŠ, where 𝑆=𝑛2(𝑇+𝑙).

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