# 7.03 Arithmetic series

Lesson

Suppose we wanted to add all of the terms in an arithmetic sequence together.  We'd be finding what's known as an arithmetic series.

#### Exploration

If the first term of an arithmetic sequence is $t_1=a$t1=a and the common difference is $d$d , then the sequence becomes:

$t_1=a$t1=a ,  $t_2=a+d$t2=a+d ,  $t_3=a+2d$t3=a+2d , $t_4=a+3d$t4=a+3d , $t_5=a+4d$t5=a+4d,  ...

Suppose we continue writing the terms all the way up to $t_{100}=a+99d$t100=a+99d , and then find a way to add up all of the one hundred terms together. The great German mathematician Carl Friedrich Gauss in 1785, at the age of 8 years old, used the following method.

$\left(t_1+t_{100}\right)+\left(t_2+t_{99}\right)+\left(t_3+t_{98}\right)+$(t1+t100)+(t2+t99)+(t3+t98)+ $...+\left(t_{48}+t_{52}\right)+\left(t_{49}+t_{51}\right)$...+(t48+t52)+(t49+t51)

Can you see why he may have done that?

Every grouped pair of terms adds up to $2a+99d$2a+99d and since there are fifty pairs, their total must be $50\times\left(2a+99d\right)$50×(2a+99d). Gauss knew there were $50$50 pairs because there are $100$100 terms. His answer could be written $\frac{100}{2}\times\left(2a+99d\right)$1002×(2a+99d) .

This led him to consider a formula for adding any number of terms of any arithmetic sequence. Writing $S_n$Sn for the sum to $n$n terms, Gauss was able to show that:

$S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$Sn=n2[2a+(n1)d]

Sum of a finite arithmetic series

The sum of a finite arithmetic series with $n$n terms, denoted $S_n$Sn , or the $n$nth partial sum of an arithmetic series can be found using one of the two related formulas:

$S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$Sn=n2[2a+(n1)d]

$S_n=\frac{n}{2}\left(a_1+a_n\right)$Sn=n2(a1+an)

#### Worked example

##### Question 1

Evaluate the sum of all the positive integers from $1$1 to $49$49.

Think/Do: With $a=1$a=1  and $d=1$d=1 we have:

$S_{49}=\frac{49}{2}\left[2\times1+\left(49-1\right)\times1\right]=1225$S49=492[2×1+(491)×1]=1225

Reflect: Note that the formula works even though there is an odd number of terms to add up.  Can you explain why?

#### Practice questions

##### QUESTION 1

Find the sum of the first $10$10 terms of the arithmetic sequence defined by $a=6$a=6 and $d=3$d=3.

##### QUESTION 2

The first term of an arithmetic sequence is $-5$5 and the $6$6th term is $-45$45.

1. If $d$d is the difference between terms, solve for $d$d.

2. Hence, find the sum of the first $14$14 terms.

##### QUESTION 3

Consider the arithmetic sequence $4$4, $-1$1, $-6$6, …

1. Write a simplified expression for the sum of the first $n$n terms.

2. Find the sum of the progression from the $19$19th to the $27$27th term, inclusive.