Arithmetic Series
How to find the value of an arithmetic series (the sum of an arithmetic sequence): formula, 3 examples, and their solutions.
Formula
Formula
A series means
the sum of the terms of a sequence.
Sn = a1 + a2 + a3 + ... + an
For an arithmetic sequence,
Sn = [n/2][2a + (n - 1)d].
Sn: The sum from a1 to an
a: First term, a1
d: Common difference
If a1 and an are given,
you can also use this formula.
Sn = [n/2][a1 + an]
Example 1
Example
Solution
a = 3
d = 5
n = 20
Then S20 = [20/2][2⋅3 + 19⋅5].
20/2 = 10
2⋅3 = 6
+19⋅5 = +95
6 + 95 = 101
10⋅101 = 1010
So S20 = 1010.
Example 2
Example
Solution
To find the number of the terms, n,
first find an.
Write -1, 3, 7, 10.
The first term, a, is -1.
-1 + 4 = 3
3 + 4 = 7
7 + 4 = 11
So write, the d, +4
between the terms.
a = -1
d = 4
So an = -1 + (n - 1)⋅4.
+(n - 1)⋅4 = +4n - 4
Multiply a Monomial and a Polynomial
-1 + 4n - 4 = 4n - 5
So an = 4n - 5.
The last term, an, is 31.
an = 4n - 5
So 4n - 5 = 31.
Move -5 to the right side.
Then 4n = 36.
Divide both sides by 4.
Then n = 9.
a = -1
d = 4
n = 9
Then S9 = [9/2][2⋅(-1) + 8⋅4].
You can also use the other formula.
n = 9
a1 = -1
an = 31
Then S9 = [9/2][-1 + 31].
2⋅(-1) = -2
+8⋅4 = +32
-2 + 32 = 30
Cancel the denominator 2
and reduce 30 to, 30/2, 15.
9⋅15 = 135
So 135 is the answer.
Example 3
Example
Solution
Sn is given.
And it says to find an.
In this case,
think of two cases:
1) n = 2, 3, 4, ... and 2) n = 1.
Case 1: n = 2, 3, 4, ...
Sn = a1 + a2 + a3 + ... + an - 1 + an
Sn - 1 = a1 + a2 + a3 + ... + an - 1
Subtract these two equations.
Then Sn - Sn - 1 = an.
So set an = Sn - Sn - 1.
This is true
when n = 2, 3, 4, ... .
Sn = n2 + 2n
So write n2 + 2n.
Write -.
Sn = n2 + 2n
So
Sn - 1 = (n - 1)2 + 2(n - 1).
So
Sn - Sn - 1
= n2 + 2n - [(n - 1)2 + 2(n - 1)].
(n - 1)2 = n2 - 2⋅n⋅1 + 12
Square of a Difference: (a - b)2
+2(n - 1) = +2n - 2
-2⋅n⋅1 = -2n
+12 = +1
Cancel n2.
Cancel +2n.
+1 - 2 = -1
-[-2n - 1] = 2n + 1
So an = 2n + 1
when n = 2, 3, 4, ... .
Next, see case 2: n = 1.
When n = 1,
an = a1.
S1 = a1
So a1 = S1.
Sn = n2 + 2n
So S1 = 12 + 2⋅1.
12 = 1
+2⋅1 = +2
1 + 2 = 3
So a1 = 3.
You got a1 = 3.
Then see if an = 2n + 1,
the an from case 1,
is true for n = 1.
In other words,
see if a1 from an = 2n + 1
is equal to 3.
a1 = 2⋅1 + 1
2⋅1 = 2
2 + 1 = 3
So a1 = 3.
This a1 is from an = 2n + 1.
a1 from S1
is 3.
And a1 from an = 2n + 1
is also 3.
a1 from an
is equal to
a1 from S1.
Then an = 2n + 1
is true for n = 1.
(If a1 from an
is not equal to
a1 from S1,
then a1 = S1
and an is not true for n = 1.)
Case 1:
an = 2n + 1
(n = 2, 3, 4, ...)
Case 2:
an = 2n + 1
(n = 1)
So an = 2n + 1.
So an = 2n + 1.