Arithmetic Series Formula

The word series implies sum. We can transform a given arithmetic sequence into an arithmetic series by adding the terms of the sequence. The example below highlights the difference between the two.

Sequence versus Series

Arithmetic Sequence (list):

\large{2,4,6,8,10,…}

Arithmetic Series (sum):

\large{2 + 4 + 6 + 8 + 10…}

Notice that in a sequence, we list the terms separated by commas while in a series, the terms are added as indicated by the plus symbols.

Therefore, an arithmetic series is simply the sum of the terms of an arithmetic sequence. More specifically, the sum of the first \large\color{red}{n} terms in an arithmetic sequence is called the partial sum. The partial sum is denoted by the symbol \large{{S_n}}.


Below is the general form of the arithmetic series formula. This works best if the first and the last terms are given in the problem.

The Arithmetic Series Formula also known as the Partial Sum Formula. Sn = n(a1+an)/2 where n=the number of terms being added, a1=the first term, an=the last term

Notes:

▶︎ The Arithmetic Series Formula is also known as the Partial Sum Formula.

▶︎ The Partial Sum Formula can be described in words as the product of the average of the first and the last terms and the total number of terms in the sum.

▶︎ The Arithmetic Sequence Formula is incorporated/embedded in the Partial Sum Formula. It is in fact the nth term or the last term \large\color{blue}{a_n} in the formula.

the nth term formula. an=a1+(n-1)d where a1=the first term, n=term position, d=common difference

▶︎ Become familiar with both the arithmetic series formula and the arithmetic sequence formula (nth term formula) because they go hand in hand when solving many problems.

\Large{{S_n} = n\left( {{{{a_1} + \,{a_n}} \over 2}} \right)}

and

\large{{a_n} = {a_1} + \left( {n - 1} \right)d}


Before we start working with examples, you may recall me mentioning that the arithmetic sequence formula is embedded in the arithmetic series formula. If we substitute and expand the nth term formula within the partial sum formula, what we will get is a new and useful form of the arithmetic series formula.

Below is the alternative formula of the arithmetic series. Consider this if the last term is not given.

Alternative Arithmetic Series Formula

Sn=n/2[2a1+(n-1)d]

where:

\large{{a_1}} is the first term

\large{{d}} is the common difference

\large{{n}} is the number of terms in the sum


Examples of Applying the Arithmetic Series Formula

Example 1: Find the sum of the first 100 natural numbers.

This is an easy problem. The purpose of this problem is to serve as an introductory example. This should help you get familiar quickly with the arithmetic series formula. Once you have a basic understanding on how to use the formula, you should be able to tackle more demanding problems as you will see later in this lesson.

Recall that the natural numbers are the counting numbers. We can write the finite arithmetic sequence as

1,2,3,4,…,100

and its related arithmetic series as

1 + 2 + 3 + 4 + … + 100

Clearly, the first term is 1, the last term is 100, and the number of terms being added is also 100.

Substitute the values into the formula then simplify to get the sum.

S sub n equals n times the average of a sub 1 and a sub n

Since {a_1} = 1, {a_{100}} = 100, and n = 100 we have

S(100)=5,050

Thus, the sum of the first 100 natural or counting numbers is 5,050.

If you want to practice more in finding the sum of the first 200, 300, 400, and 500 natural numbers, you can use the list of partial sums of natural numbers up to 1,000 that I created as an answer key.


Example 2: Find the partial sum of the given arithmetic series.

\large{7 + 12 + 17 + 22 + … + 187}

If this is your first time solving this type of problem, you may find this a bit overwhelming. Not that it is difficult but because the values that you need are not explicitly given. This can throw you off because you don’t even know how to get started. However, if you have a strategy from the beginning you will realize this problem is not that bad.

What we need to do is to examine the given series. Identify the values that are pertinent and useful to us. Sometimes by doing it this way, the next logical step will be revealed to us.

a diagram showing the first term as 7, the last term as 187, and the common difference as 5

So this is the information we gathered from the series. The first term is 7. Since 12-7=5, 17-12=5, and 22-17=5, then the common difference is 5. The last term is 187. That means the number of terms \large\color{red}n being added in the series is missing.

\large{a_1} = 7

\large{d=5}

\large{a_n} = 187

\large{n = \,?}

I hope at this time, you can agree with me that we have no other option but to use the nth term formula to find \large\color{red}n. Once we find the value for \large{n}, we will substitute that into the arithmetic series formula together with the first and last terms to find the sum of the given arithmetic series.

an is given, a1 is given, d is easily computed.

Now, let’s find \large{n} using the nth term formula.

n=37

Finally, we have all the values that we need to calculate the sum of the given series which are \large{n=37}, \large{a_1} = 7, and \large{a_n} = 187.

37th partial sum equals 3,589

Example 3: Find the sum of the first \large{51} terms of the arithmetic sequence.

\large{12\,,\,19\,,\,26\,,\,33\,,...}

The strategy here is similar to Example 2. Instead of finding the number of terms \large\color{red}n, we will use the nth term formula to find the 51st term. Then, we use the the arithmetic series formula to calculate the sum of the first 51 terms of the sequence.

an of nth term formula substituted to an of sn partial sum formula

So what value can we extract from the given problem?

Well, the number of terms \large\color{red}n to be added is explicitly given in the problem which is n=51.

Now, from the arithmetic sequence, the first term and common difference are easily identifiable. The first term is obviously 12 while the common difference is 7 since 19 - 12 = 7, 26 - 19 = 7, and 33 - 26 = 7.

first term is 12 while common difference is 7

So here is the information we have gathered. It means the nth term is what we are looking for.

\large{a_1}=12

\large{n=51}

\large{d=7}

\large{a_n}=\,?

Plug in the values into the nth term formula then simplify to get the 51st term.

the 51st term is 362

We can finally find the sum of the first 51 terms because we know the number of terms n=51, the first term {a_1}=7, and the last term {a_n}=362.

the sum of the first 51 terms is 9,537

Example 4: The 10th term of an arithmetic sequence is 17 and the 30th term is -63. What is the 50th partial sum \large{S_{50}} of the arithmetic sequence?

Here’s the big picture. In order to find the 50th partial sum, we will need to know the first term \large{a_1}, and the last term \large{a_n} which is the same as the 50th term. Obviously, there will be 50 terms in the series because we are summing up the terms from the first up to the 50th term.

To find the first term \large{a_1}, we will use the nth term formula together with the given information in the problem to generate a system of equations where the unknown variables are the first term \large{a_1} and the common difference d.

\large{{a_n} = {a_1} + \left( {n - 1} \right)d}

Therefore, we have

  • 10th term is 17
equation 1 is 17=a1+9d
  • 30th term is -63
equation 2 is -63=a1+29d

Now, this is the system of equations that we are going to solve. We can find the values of the first term \large{a_1} and the common difference \large{d}.

equation 1 is 17=a+9d, equation 2 is -63=a+29d

We will solve this system of equations using the Elimination Method. We subtract equation #2 from equation #1 to get rid of \large{a_1}, thus isolating \large{d}.

17=a+9d minus -63=a+29d

This gives us

d=-4

Since we already know the value of the common difference \large{d}, we can easily solve for the first term \large{a_1}. Choose any of the two equations, equation #1 or equation #2, substitute the value of \large{d} then solve for \large{a_1}. We will choose equation #1 because it is a much simpler equation to deal with.

a1=53

By knowing the first term, and the common difference of a sequence, we can put together the formula that can determine any term in the sequence.

an=53+(n-1)(-4)

Using the formula that we have come up with, we can now find the 50th term \large{{a_{50}}} in the sequence.

a50=-143

Finally, we have all that we need to compute for the 50th partial sum using the arithmetic series formula.

a1=53, a50=-143, n=50

Substitute the values into the formula then simplify.

S50=-2,250

Example 5: The 10th term of an arithmetic sequence is 23 while its 12th partial sum is 192. Find the sum of the first 40 terms of the sequence.

To find the first 40 terms of the arithmetic sequence, we will use the main arithmetic series formula. However, we need to supply the missing values in the formula, namely the first term \large{{{a_1}}} and the last term \large{{{a_n}}}. The number of terms to be added \large{n} is already given which is 40.

Sn = n[{a1+an)/2]

Now, let’s construct a system of equations wherein the unknown variables are the first term \large{a{}_1} and the common difference \large{d}.

The first equation comes from the given information that \large{{a_{10}} = 23}. Plug the values into the nth term formula.

23=a1+9d

The second equation comes from the given information that \large{{S_{12}} = 192}. Plug the values into the alternative arithmetic series formula.

12a1 + 66d = 192

This is the system of equations that we are going to solve by the Elimination Method.

equation 1 is 192=12a1+66d while equation 2 is 23a1 + 9d

Multiply equation #1 by -12.

-12 (23a1+9d)

Then add it to equation #2. We get {d=2}.

d equals 2

After solving for the value of \large{d}, we can now solve for the value of \large{a{}_1}. Pick any of the two equations, substitute the value of \large{d} then solve for \large{a{}_1}. We will use equation #1 because it is a simpler equation.

a sub 1 equals 5

Since we know already the values of \large{a{}_1} and \large{d}, we are now ready to write the general term of the sequence which can find any term in the sequence.

a sub n equals 5 plus n minus 1 times 2

To find the 40th term, we have

a sub 40 equals 83

Finally, we have all the required values as shown below to calculate the 40th partial sum.

\large{n=40}

\large{{a_1} = 5}

\large{{{a_n} = 83}}

Plug in the values into the arithmetic series formula then simplify.

S sub 40 equals 1760

You might also be interested in:

Derivation of the Arithmetic Series Formula

Arithmetic Series Formula Practice Problems

Arithmetic Sequence Formula

Arithmetic Sequence Formula Practice Problems

Geometric Sequence Formula