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Wize High School Algebra II Textbook (Common Core) > Sequences & Series

Arithmetic Series

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Arithmetic Series

What is a Series?

  • A series is the sum of the terms of a sequence, usually denoted SS.
  • An arithmetic series is the sum of the terms of an arithmetic sequence.
  • A partial sum of a series is the sum of the first n\bm n terms of the sequence, denoted SnS_n.
Example

Consider the arithmetic sequence 1,2,3,4, …1,2,3,4,\ \dots.
We can find the associated arithmetic series by replacing the commas with ++ signs:
S=1+2+3+4+…S=1+2+3+4+\dots
The partial sum of the first three terms is:
S3=1+2+3=6S_3 = 1+2+3=6
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Gauss's Method for Arithmetic Series

As a child, famous mathematician Gauss was asked to add the numbers from 1 to 100. This is how he did it:

Define the arithmetic series S=1+2+3+⋯+99+100S=1+2+3+ \dots + 99 + 100.

Let's write the series in two different ways and add up corresponding pairs:
  • S=1+2+⋯+99+100S=\colorOne{1+2+\dots+99+100}
  • S=100+99+⋯+2+1S=\colorTwo{100+99+\dots+2+1}
If we add the series to itself in this way, what do you notice about the sum of each pair?


During this process, we added S+SS+S. The sum of each pair of terms is always 101\bm{101}, and this happens 100 times:
2S=100×101S=100×1012S=5050\begin{aligned} \bm{2S} &= 100 \times \bm{101}\\[0.5em] S &= \dfrac{100 \times 101}{2}\\[1em] S &= \boxed{5050} \end{aligned}
General Formula

The partial sum SnS_n of an arithmetic series can be calculated with the formula:
Sn=n(t1+tn)2\boxed{\quad S_n = \dfrac{n(t_1 + t_n)}{2} \quad}
  • nn is the number of terms to add
  • t1t_1 is the first term, and tnt_n is the last term we want to include
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Another Formula for Arithmetic Series

If you know the first term, aa, and the common difference, dd, we have another formula for arithmetic series:
Sn=n[2a+(n−1)d]2\boxed{\quad S_n = \dfrac{n[2a + (n-1)d]}{2} \quad}

Example

Compute S9S_9 for the sequence 4,7,10,13, …4,7,10,13,\ \dots

The first term is a=4a=4
The common difference is d=3d=3

S9=9[2a+(9−1)d]2=9[2(4)+(8)(3)]2=9(8+24)2=9(32)2=144\begin{aligned} S_9 &= \dfrac{9[2a + (9-1)d]}{2}\\[1em] &= \dfrac{9[2(4) + (8)(3)]}{2}\\[1em] &= \dfrac{9(8+24)}{2}\\[1em] &= \dfrac{9(32)}{2}\\[1em] &= \boxed{144}\\[1em] \end{aligned}
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Example: Arithmetic Series


A charity is celebrating its 40th anniversary and poses a challenge to celebrate the occasion.

The first donor is to give a single $5 bill, and each subsequent donor is asked to give one more $5 bill than the previous donor. This will continue up until the 40th donation.

How much money will be raised by the end of this challenge?

The series is given by S=1(5)+2(5)+3(5)+⋯+40(5)=5+10+15+⋯+200\begin{aligned} S &= 1(5) + 2(5) + 3(5)+ \dots+ 40(5)\\ &= 5 + 10 + 15+ \dots+ 200 \end{aligned}.

Since this is an arithmetic series, we can use either of the formulas.

Method 1

There are n=40n=40 terms, the first term is t1=5t_1=5 and the last term is t40=200t_{40}=200.

Sn=n(t1+tn)2S40=n(t1+t40)2=40(5+200)2=$ 4100\begin{aligned} S_{n} &= \dfrac{n(t_1 + t_{n})}{2}\\[1em] S_{40} &= \dfrac{n(t_1 + t_{40})}{2}\\[1em] &= \dfrac{40(5 + 200)}{2}\\[1em] &= \boxed{\$\,4100}\\[1em] \end{aligned}
Method 2

There are n=40n=40 terms, the first term is a=5a=5, and the common difference is d=5d=5.

Sn=n[2a+(n−1)d]2S40=40[2(5)+(40−1)(5)]2=40[10+195]2=$ 4100\begin{aligned} S_n &= \dfrac{n[2a + (n-1)d]}{2} \\[1em] S_{40} &= \dfrac{40[2(5) + (40-1)(5)]}{2} \\[1em] &= \dfrac{40[10+195]}{2} \\[1em] &= \boxed{\$\,4100} \\[1em] \end{aligned}

Practice: Arithmetic Series

Match each arithmetic series with the proper values.
A.
n=7n=7
B.
S6=45S_6=45
C.
d=9d=9
D.
a=t1=3a=t_1=3
4+10+16+⋯+404+10+16+\dots+40
1+10+19+⋯+371+10+19+\dots+37
3+5+7+⋯+213+5+7+\dots+21
5+6+7+⋯+105+6+7+\dots+10

Practice: Arithmetic Series

What is the sum of the first 10 terms of an arithmetic series with t3=11t_3=11 and t7=27t_7=27?

Practice: Arithmetic Series



A space shuttle is launching off from Earth towards the Moon. Every second, the spacecraft travels 100m farther than it did the previous second. If it starts at rest for the first second, how far has the space shuttle travelled after 30 seconds?