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.

If the first term of an arithmetic sequence is $t_1=a$`t`1=`a` and the common difference is $d$`d` , then the sequence becomes:

$t_1=a$`t`1=`a` , $t_2=a+d$`t`2=`a`+`d` , $t_3=a+2d$`t`3=`a`+2`d` , $t_4=a+3d$`t`4=`a`+3`d` , $t_5=a+4d$`t`5=`a`+4`d`, ...

Suppose we continue writing the terms all the way up to $t_{100}=a+99d$`t`100=`a`+99`d` , 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.

Instead of adding the terms from left to right, he added them in this order:

$\left(t_1+t_{100}\right)+\left(t_2+t_{99}\right)+\left(t_3+t_{98}\right)+$(`t`1+`t`100)+(`t`2+`t`99)+(`t`3+`t`98)+ $...+\left(t_{48}+t_{52}\right)+\left(t_{49}+t_{51}\right)$...+(`t`48+`t`52)+(`t`49+`t`51)

Can you see why he may have done that?

Every grouped pair of terms adds up to $2a+99d$2`a`+99`d` and since there are fifty pairs, their total must be $50\times\left(2a+99d\right)$50×(2`a`+99`d`). 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×(2`a`+99`d`) .

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

$S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$`S``n`=`n`2[2`a`+(`n`−1)`d`]

Sum of a finite arithmetic series

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

$S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$`S``n`=`n`2[2`a`+(`n`−1)`d`]

$S_n=\frac{n}{2}\left(a_1+a_n\right)$`S``n`=`n`2(`a`1+`a``n`)

**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:

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

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

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

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

If $d$

`d`is the difference between terms, solve for $d$`d`.Hence, find the sum of the first $14$14 terms.

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

Write a simplified expression for the sum of the first $n$

`n`terms.Find the sum of the progression from the $19$19th to the $27$27th term, inclusive.