Remember that an arithmetic sequence is a set of numbers, such that the difference between any term and the previous term is a constant number, dd, called the constant difference:
a
n
=
a
1
+
d
(
n
-
1
)
a
n
=
a
1
+
d
(
n
-
1
)
(14)where
- nn is the index of the sequence;
- anan is the nthnth-term of the sequence;
- a1a1 is the first term;
- dd is the common difference.
When we sum a finite number of terms in an arithmetic sequence, we get a finite arithmetic series.
The simplest arithmetic sequence is when a1=1a1=1 and d=0d=0 in the general form Equation 14; in other words all the terms in the sequence are 1:
a
i
=
a
1
+
d
(
i
-
1
)
=
1
+
0
·
(
i
-
1
)
=
1
{
a
i
}
=
{
1
;
1
;
1
;
1
;
1
;
...
}
a
i
=
a
1
+
d
(
i
-
1
)
=
1
+
0
·
(
i
-
1
)
=
1
{
a
i
}
=
{
1
;
1
;
1
;
1
;
1
;
...
}
(15)If we wish to sum this sequence from i=1i=1 to any positive integer nn, we would write
∑
i
=
1
n
a
i
=
∑
i
=
1
n
1
=
1
+
1
+
1
+
...
+
1
(
n
times
)
∑
i
=
1
n
a
i
=
∑
i
=
1
n
1
=
1
+
1
+
1
+
...
+
1
(
n
times
)
(16)Since all the terms are equal to 1, it means that if we sum to nn we will be adding nn-number of 1's together, which is simply equal to nn:
∑
i
=
1
n
1
=
n
∑
i
=
1
n
1
=
n
(17)Another simple arithmetic sequence is when a1=1a1=1 and d=1d=1, which is the sequence of positive integers:
a
i
=
a
1
+
d
(
i
-
1
)
=
1
+
1
·
(
i
-
1
)
=
i
{
a
i
}
=
{
1
;
2
;
3
;
4
;
5
;
...
}
a
i
=
a
1
+
d
(
i
-
1
)
=
1
+
1
·
(
i
-
1
)
=
i
{
a
i
}
=
{
1
;
2
;
3
;
4
;
5
;
...
}
(18)If we wish to sum this sequence from i=1i=1 to any positive integer nn, we would write
∑
i
=
1
n
i
=
1
+
2
+
3
+
...
+
n
∑
i
=
1
n
i
=
1
+
2
+
3
+
...
+
n
(19)This is an equation with a very important solution as it gives the answer to the sum of positive integers.
Mathematician, Karl Friedrich Gauss, discovered this proof when he was only 8 years old. His teacher had decided to give his class a problem which would distract them for the entire day by asking them to add all the numbers from 1 to 100. Young Karl realised how to do this almost instantaneously and shocked the teacher with the correct answer, 5050.
We first write SnSn as a sum of terms in ascending order:
S
n
=
1
+
2
+
...
+
(
n
-
1
)
+
n
S
n
=
1
+
2
+
...
+
(
n
-
1
)
+
n
(20)We then write the same sum but with the terms in descending order:
S
n
=
n
+
(
n
-
1
)
+
...
+
2
+
1
S
n
=
n
+
(
n
-
1
)
+
...
+
2
+
1
(21)We then add corresponding pairs of terms from equations Equation 20 and Equation 21, and we find that the sum for each pair is the same, (n+1)(n+1):
2
S
n
=
(
n
+
1
)
+
(
n
+
1
)
+
...
+
(
n
+
1
)
+
(
n
+
1
)
2
S
n
=
(
n
+
1
)
+
(
n
+
1
)
+
...
+
(
n
+
1
)
+
(
n
+
1
)
(22)We then have nn-number of (n+1)(n+1)-terms, and by simplifying we arrive at the final result:
2
S
n
=
n
(
n
+
1
)
S
n
=
n
2
(
n
+
1
)
2
S
n
=
n
(
n
+
1
)
S
n
=
n
2
(
n
+
1
)
(23)
S
n
=
∑
i
=
1
n
i
=
n
2
(
n
+
1
)
S
n
=
∑
i
=
1
n
i
=
n
2
(
n
+
1
)
(24)Note that this is an example of a quadratic sequence.
If we wish to sum any arithmetic sequence, there is no need to work it out term-for-term. We will now determine the general formula to evaluate a finite arithmetic series. We start with the general formula for an arithmetic sequence and sum it from i=1i=1 to any positive integer nn:
∑
i
=
1
n
a
i
=
∑
i
=
1
n
[
a
1
+
d
(
i
-
1
)
]
=
∑
i
=
1
n
(
a
1
+
d
i
-
d
)
=
∑
i
=
1
n
[
(
a
1
-
d
)
+
d
i
]
=
∑
i
=
1
n
(
a
1
-
d
)
+
∑
i
=
1
n
(
d
i
)
=
∑
i
=
1
n
(
a
1
-
d
)
+
d
∑
i
=
1
n
i
=
(
a
1
-
d
)
n
+
d
n
2
(
n
+
1
)
=
n
2
(
2
a
1
-
2
d
+
d
n
+
d
)
=
n
2
(
2
a
1
+
d
n
-
d
)
=
n
2
[
2
a
1
+
d
(
n
-
1
)
]
∑
i
=
1
n
a
i
=
∑
i
=
1
n
[
a
1
+
d
(
i
-
1
)
]
=
∑
i
=
1
n
(
a
1
+
d
i
-
d
)
=
∑
i
=
1
n
[
(
a
1
-
d
)
+
d
i
]
=
∑
i
=
1
n
(
a
1
-
d
)
+
∑
i
=
1
n
(
d
i
)
=
∑
i
=
1
n
(
a
1
-
d
)
+
d
∑
i
=
1
n
i
=
(
a
1
-
d
)
n
+
d
n
2
(
n
+
1
)
=
n
2
(
2
a
1
-
2
d
+
d
n
+
d
)
=
n
2
(
2
a
1
+
d
n
-
d
)
=
n
2
[
2
a
1
+
d
(
n
-
1
)
]
(25)So, the general formula for determining an arithmetic series is given by
S
n
=
∑
i
=
1
n
[
a
1
+
d
(
i
-
1
)
]
=
n
2
[
2
a
1
+
d
(
n
-
1
)
]
S
n
=
∑
i
=
1
n
[
a
1
+
d
(
i
-
1
)
]
=
n
2
[
2
a
1
+
d
(
n
-
1
)
]
(26)For example, if we wish to know the series S20S20 for the arithmetic sequence ai=3+7(i-1)ai=3+7(i-1), we could either calculate each term individually and sum them:
S
20
=
∑
i
=
1
20
[
3
+
7
(
i
-
1
)
]
=
3
+
10
+
17
+
24
+
31
+
38
+
45
+
52
+
59
+
66
+
73
+
80
+
87
+
94
+
101
+
108
+
115
+
122
+
129
+
136
=
1390
S
20
=
∑
i
=
1
20
[
3
+
7
(
i
-
1
)
]
=
3
+
10
+
17
+
24
+
31
+
38
+
45
+
52
+
59
+
66
+
73
+
80
+
87
+
94
+
101
+
108
+
115
+
122
+
129
+
136
=
1390
(27)or, more sensibly, we could use equation Equation 26 noting that a1=3a1=3, d=7d=7 and n=20n=20 so that
S
20
=
∑
i
=
1
20
[
3
+
7
(
i
-
1
)
]
=
20
2
[
2
·
3
+
7
(
20
-
1
)
]
=
1390
S
20
=
∑
i
=
1
20
[
3
+
7
(
i
-
1
)
]
=
20
2
[
2
·
3
+
7
(
20
-
1
)
]
=
1390
(28)This example demonstrates how useful equation Equation 26 is.
- The sum to nn terms of an arithmetic series is Sn=n2(7n+15)Sn=n2(7n+15).
- How many terms of the series must be added to give a sum of 425?
- Determine the 6th term of the series.
- The sum of an arithmetic series is 100 times its first term, while the last term is 9 times the first term. Calculate the number of terms in the series if the first term is not equal to zero.
- The common difference of an arithmetic series is 3. Calculate the values of nn for which the nthnth term of the series is 93, and the sum of the first nn terms is 975.
- The sum of nn terms of an arithmetic series is 5n2-11n5n2-11n for all values of nn. Determine the common difference.
- The sum of an arithmetic series is 100 times the value of its first term, while the last term is 9 times the first term. Calculate the number of terms in the series if the first term is not equal to zero.
- The third term of an arithmetic sequence is -7 and the 7th term is 9. Determine the sum of the first 51 terms of the sequence.
- Calculate the sum of the arithmetic series 4+7+10+⋯+9014+7+10+⋯+901.
- The common difference of an arithmetic series is 3. Calculate the values of nn for which the n th n th term of the series is 93 and the sum of the first nn terms is 975.
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