Arithmetic Series
Understand the difference
between a sequence and a series
Proving the nth term rule
Proving the formula to find the
sum of an arithmetic series
Consider the infinite sequence
4,7,10,13,….
If the terms of the sequence are
added this becomes a finite series
4+7+10+13
In an arithmetic series the
difference between the terms is
constant.
The difference is called the common
difference
An arithmetic series is also known as
an arithmetic progression (AP)
Using the sequence 4, 7, 10, 13…
a=1st term of the sequence
d=common difference
n
1
2
3
4
3n+1
4
7
10
13
a
a+d
a+2d
a+3d
So the nth term would be….
a + (n-1)d
Proof the the sum of an Arithmetic Series
n
1
3n+1 4
2
7
3
10
4 ….. 19
13 ….. 58
20
61
Call the sum of the terms Sn
Sn= 4 + 7 + 10 + 13 + ….. + 58 + 61
Reverse the order
Sn= 61+58 + 55+ 52 + ….. + 4 + 7
Add the two series together
2Sn = 65 + 65 + 65 + 65 + ….. + 65 + 65
2Sn = 65x 20 (because there are 20 terms)
2Sn = 1300
Sn = 650 (divide by 2)
Proof the the sum of an Arithmetic Series
a=first term, d=common difference, L=last term
n
1 2
3
4
a a+d a+2d a+3d
….. n-1 n
….. L-d L
Sum the first n terms then reverse the order
Sn= a + (a+d) + (a+2d) + (a+3d) + ….. + (L-2d) + (L-d) + L
Sn= L + (L-d) + (L-2d) + (L-3d) + ….. +(a+2d) + (a+d)+ a
Add the two series together
2Sn= (a+L)+(a+L)+ (a+L) + (a+L) + ….. + (a+L) + (a+L)+(a+L)
2Sn = n(a+L) (because there are n terms)
Sn = n(a+L)
2
Nearly there!!
Proof the the sum of an Arithmetic Series
a=first term, d=common difference, L=last term
Sn = n(a+L)
2
L (the last term) is also the nth term which we
know has the formula a+(n-1)d so if we
substitute for L in the formula above we get….
Sn = n[a+a+(n-1)d]
2
Sn = n[2a+(n-1)d]
2
You need to learn this formula
EXAMPLE 1
Find the sum of the first 30 terms in the
series 3+9+15+…
a=3, d=6, n=30
Using the formula
Sn = n[2a+(n-1)d]
2
Sn = 30[2x3+(30-1)6]
2
Sn = 15[6+(29x6)]
Sn = 15x180 = 2700
EXAMPLE 2
a)Find the nth term of the arithmetic series 7+11+15+..
b)Which term of the sequence is equal to 51?
c)Hence find 7+11+15+…+51
a) a=7, d=4 so the nth term is 4n+3
b) 4n+3= 51
4n = 48 (subtract 3)
n = 12 (divide by 4)
c) Using the formula
Sn = n[2a+(n-1)d] a=7, d=4 and n=12
2
Sn = 12[2x7+(12-1)4]
2
Sn = 6[14+(11x4)]
Sn = 6x58 = 348