An arithmetic sequenceA sequence of numbers where each successive number is the sum of the previous number and some constant d., or arithmetic progressionUsed when referring to an arithmetic sequence., is a sequence of numbers where each successive number is the sum of the previous number and some constant d.
an=an−1+d Arithmetic Sequence
And because an−an−1=d, the constant d is called the common differenceThe constant d that is obtained from subtracting any two successive terms of an arithmetic sequence; an−an−1=d.. For example, the sequence of positive odd integers is an arithmetic sequence,
1,3,5,7,9,…
Here a1=1 and the difference between any two successive terms is 2. We can construct the general term an=an−1+2 where,
a1=1a2=a1+2=1+2=3a3=a2+2=3+2=5a4=a3+2=5+2=7a5=a4+2=7+2=9⋮
In general, given the first term a1 of an arithmetic sequence and its common difference d, we can write the following:
a2=a1+da3=a2+d=(a1+d)+d=a1+2da4=a3+d=(a1+2d)+d=a1+3da5=a4+d=(a1+3d)+d=a1+4d⋮
From this we see that any arithmetic sequence can be written in terms of its first element, common difference, and index as follows:
an=a1+(n−1)d Arithmetic Sequence
In fact, any general term that is linear in n defines an arithmetic sequence.
Example 1
Find an equation for the general term of the given arithmetic sequence and use it to calculate its 100th term: 7,10,13,16,19,…
Solution:
Begin by finding the common difference,
d=10−7=3
Note that the difference between any two successive terms is 3. The sequence is indeed an arithmetic progression where a1=7 and d=3.
an=a1+(n−1)d=7+(n−1)⋅3=7+3n−3=3n+4
Therefore, we can write the general term an=3n+4. Take a minute to verify that this equation describes the given sequence. Use this equation to find the 100th term:
a100=3(100)+4=304
Answer: an=3n+4; a100=304
The common difference of an arithmetic sequence may be negative.
Example 2
Find an equation for the general term of the given arithmetic sequence and use it to calculate its 75th term: 6,4,2,0,−2,…
Solution:
Begin by finding the common difference,
d=4−6=−2
Next find the formula for the general term, here a1=6 and d=−2.
an=a1+(n−1)d=6+(n−1)⋅(−2)=6−2n+2=8−2n
Therefore, an=8−2n and the 75th term can be calculated as follows:
a75=8−2(75)=8−150=−142
Answer: an=8−2n; a100=−142
The terms between given terms of an arithmetic sequence are called arithmetic meansThe terms between given terms of an arithmetic sequence..
Example 3
Find all terms in between a1=−8 and a7=10 of an arithmetic sequence. In other words, find all arithmetic means between the 1st and 7th terms.
Solution:
Begin by finding the common difference d. In this case, we are given the first and seventh term:
an=a1+(n−1)d Use n=7.a7=a1+(7−1)da7=a1+6d
Substitute a1=−8 and a7=10 into the above equation and then solve for the common difference d.
10=−8+6d18=6d3=d
Next, use the first term a1=−8 and the common difference d=3 to find an equation for the nth term of the sequence.
an=−8+(n−1)⋅3=−8+3n−3=−11+3n
With an=3n−11, where n is a positive integer, find the missing terms.
a1=3(1)−11=3−11=−8a2=3(2)−11=6−11=−5a3=3(3)−11=9−11=−2a4=3(4)−11=12−11=1a5=3(5)−11=15−11=4a6=3(6)−11=18−11=7 } arithmetic meansa7=3(7)−11=21−11=10
Answer: −5, −2, 1, 4, 7
In some cases, the first term of an arithmetic sequence may not be given.
Example 4
Find the general term of an arithmetic sequence where a3=−1 and a10=48.
Solution:
To determine a formula for the general term we need a1 and d. A linear system with these as variables can be formed using the given information and an=a1+(n−1)d:
{a3=a1+(3−1)da10=a1+(10−1)d⇒ {−1=a1+2d48=a1+9d Use a3=−1. Use a10=48.
Eliminate a1 by multiplying the first equation by −1 and add the result to the second equation.
{−1=a1+2d48=a1+9d ⇒×(−1) + {1=−a1−2d48= a1+ 9d¯ 49=7d7=d
Substitute d=7 into −1=a1+2d to find a1.
−1=a1+2(7)−1=a1+14−15=a1
Next, use the first term a1=−15 and the common difference d=7 to find a formula for the general term.
an=a1+(n−1)d=−15+(n−1)⋅7=−15+7n−7=−22+7n
Answer: an=7n−22
Try this! Find an equation for the general term of the given arithmetic sequence and use it to calculate its 100th term: 32,2,52,3,72,…
If a series is arithmetic the sum of the first n terms, denoted Sn , there are ways to find its sum without actually adding all of the terms.
To find the sum of the first n terms of an arithmetic series use the formula, n terms of an arithmetic sequence use the formula,
Sn=n(a1 + an)2 ,
where n is the number of terms, a1 is the first term and an is the last term.
The series 3+6+9+12+⋯+30 can be expressed as sigma notation ∑n=1103n . This expression is read as the sum of 3n as n goes from 1 to 10
Example 1:
Find the sum of the first 20 terms of the arithmetic series if a1=5 and a20=62 .
S20=20(5 + 62)2S20=670
Example 2:
Find the sum of the first 40 terms of the arithmetic sequence
2,5,8,11,14,⋯
First find the 40 th term:
a40=a1+(n−1)d =2+39(3)=119
Then find the sum:
Sn=n(a1+an)2S40=40(2 + 119)2=2420
Example 3:
Find the sum:
∑k=150(3k+2)
First find a1 and a50 :
a1=3(1)+2=5a20=3(50)+2=152
Then find the sum:
Sk=k(a1 + ak)2S50=50(5 + 152)2=3925