Sum of Arithmetic Sequence – Formulas and Examples

Arithmetic sequences are formed by adding the common difference to each previous term. We can calculate the sum of arithmetic sequences by applying a formula with the values of the first term and the last term, in addition to the number of terms.

Here, we will learn how to obtain the sum of an arithmetic sequence. We will explore the formulas that we can use and apply them to solve some examples.

ALGEBRA
Formula for the sum of an arithmetic sequence

Relevante para

Aprender a encontrar la suma de una progresión aritmética.

See formulas

ALGEBRA
Formula for the sum of an arithmetic sequence

Relevant for

Learning to find the sum of an arithmetic sequence.

See formulas

Formulas for the sum of an arithmetic sequence

Arithmetic sequences are sequences in which their terms are formed from the previous term by adding a certain number called the common difference.

The sum of the first $latex n$ terms of an arithmetic sequence can be found with the following formula

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

where,

  • $latex a$ is the first term of the sequence.
  • $latex l$ is the last term.
  • $latex n $ is the number of terms.
Formula for the sum of an arithmetic sequence

Also, remembering that any term of an arithmetic sequence can be found using the formula $latex a_{n}=a+(n-1)d$, we can write the formula for the sum as follows:

$$S_{n}=\frac{n}{2}[2a+(n-1)d]$$

where,

  • $latex a$ is the first term.
  • $latex d$ is the common difference.
  • $latex n $ is the number of terms.

Proof of the formula for the sum of arithmetic sequences

Recall that each term of an arithmetic sequence is obtained by adding the common difference, d, to the previous term. Therefore, we can write the following:

$$S_{n}=a+[a+d]+…+[a+(n-1)d]$$

This is equation [1]. Now, we write the terms in reverse order, that is, from right to left.

$$S_{n}=[a+(n-1)d]+[a+(n-2)d]+…+a$$

This is equation [2]. If we add both equations, we can obtain the value of $latex 2S_{n}$:

$$2S_{n}=(a+[a+(n-1)d])+((a+d)+[a+(n-2)d])+…+([a+(n-1)d]+a)$$

$$2S_{n}=[2a+(n-1)d]+[2a+(n-1)d]+…+[2a+(n-1)d]$$

We can see that the terms obtained are equal, so the sum is equal to one of the terms multiplied by n (total number of terms).

$$2S_{n}=n[2a+(n-1)d]$$

Finally, we divide the entire equation by 2 to find $latex S_{n}$:

$$S_{n}=\frac{n}{2}[2a+(n-1)d]$$


Solved examples of sum of arithmetic sequences

EXAMPLE 1

The first term of an arithmetic sequence is 4 and the 10th term is 31. What is the sum of the first 10 terms?

We have the following information:

  • $latex a=4$
  • $latex l=31$
  • $latex n=10$

Now, we use the formula for the sum of arithmetic progressions with the given values:

$$S_{n}=\frac{n}{2}[a+l]$$

$$S_{10}=\frac{10}{2}[4+31]$$

$$S_{10}=5[35]$$

$$S_{10}=175$$

EXAMPLE 2

Find the sum of the first 30 terms of an arithmetic sequence, where the first term equals 5 and the 30th term equals 179.

Similar to the previous example, we can observe the following values:

  • $latex a=5$
  • $latex l=179$
  • $latex n=30$

Now, we use the formula for the sum of arithmetic progressions with the given information:

$$S_{n}=\frac{n}{2}[a+l]$$

$$S_{30}=\frac{30}{2}[5+179]$$

$$S_{30}=15[184]$$

$$S_{30}=2760$$

EXAMPLE 3

The first four terms of an arithmetic sequence are 5, 9, 13, and 17. Find the sum of the first 20 terms.

We can find the common difference of the sequence by subtracting a term by its previous term. For example, 17-13=4. Then, we have:

  • $latex a=5$
  • $latex d=4$
  • $latex n=20$

In this case, we don’t know the last term of the required sum, that is, the 20th term. Thus, we can use the second formula with the given information:

$$S_{n}=\frac{n}{2}[2a+(n-1)d]$$

$$S_{20}=\frac{20}{2}[2(5)+(20-1)4]$$

$$=10[10+(19)4]$$

$latex =10[10+76]$

$latex =10(86)$

$latex S_{20}=860$

EXAMPLE 4

Find the sum of the first 24 terms of an arithmetic sequence that starts with the terms -2, -5, -8, -11, …

The common difference of the progression is -11-(-8)=-3. Then, we have the following:

  • $latex a=-2$
  • $latex d=-3$
  • $latex n=24$

Similar to the previous example, we can use the second formula for the sum of an arithmetic progression with the given values:

$$S_{n}=\frac{n}{2}[2a+(n-1)d]$$

$$S_{24}=\frac{24}{2}[2(-2)+(24-1)(-3)]$$

$$=12[-4+(23)(-3)]$$

$latex =12[-4-69]$

$latex =12(-73)$

$latex S_{24}=-876$

EXAMPLE 5

Find the sum of the following arithmetic sequence:

$$6+8+10+…+30$$

The common difference is equal to 10-8=2. Then, we have the following:

  • $latex a=6$
  • $latex d=2$
  • $latex l=30$

However, to find the sum, we have to know the number of terms, n. We can achieve this using the formula for the nth term:

$latex a_{n}=a+(n-1)d$

$latex 30=6+(n-1)2$

$latex 24=(n-1)2$

$latex 12=n-1$

$latex n=13$

Now, we find the sum with these values:

$$S_{n}=\frac{n}{2}[a+l]$$

$$S_{13}=\frac{13}{2}[6+30]$$

$$S_{13}=\frac{13}{2}[36]$$

$$S_{13}=234$$

EXAMPLE 6

The first term of an arithmetic sequence is 2 and the nth term is 32. If the sum of the first n terms is 357, find the value of n.

Since the nth term is 32, we have:

$latex a+(n-1)d=32$

In addition, we also know that $latex a =2$, so we have:

$latex 2+(n-1)d=32$

$latex (n-1)d=30~~[1]$

Since the sum of the first n terms is 357, we have:

$$ \frac{n}{2}[2a+(n-1)d]=357$$

We also know that $latex a=2$, so we have:

$$ \frac{n}{2}[2(2)+(n-1)d]=357$$

$$ \frac{n}{2}[4+(n-1)d]=357~~[2]$$

Substituting equation [1] into equation [2], we have:

$latex n(4+30)=714$

$latex 34n=714$

$latex n=21$

Explore additional examples of the sum of an arithmetic sequence by visiting this article: Sum of Arithmetic Sequence – Examples and Practice Problems.


Sum of arithmetic sequences – Practice problems

Sum of arithmetic sequences quiz
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Find the sum of the following arithmetic sequence: $$9+13+17+…+41$$

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$latex S=$

See also

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