# Sum_of_an_Arithmetic_Series

ARITHMETIC SEQUENCES
AND SERIES
Week Commencing Monday 5th October
Learning Intention:
•
To be able to find the sum of an
arithmetic series
Contents:
1. Sum of an Arithmetic Series Formula
2. Sum Formula Proof
3. Using the Sum Formula
4. Real-Life Problems
5. Assignment 4
ARITHMETIC SEQUENCES
AND SERIES
Sum of an Arithmetic Series
The sum of the first n terms of a series is
generally denoted by Sn.
For arithmetic series there is a formula to work
out the sum of the first n terms:
Sn 
n
2
2a  (n -1)d
where n is the number of terms
a is the first term
d is the common difference
ARITHMETIC SEQUENCES
AND SERIES
Proof for Sum of an Arithmetic Series
You need to be able to reproduce this proof for
the Core 1 examination.
Sn  a  (a  d)  (a  2d)  ...  (a  (n -1)d)
This can be written in reverse :
Sn  (a  (n -1)d)  (a  (n - 2)d)  ...  (a  d)  a
Adding the two sums together we get :
2Sn  (2a  (n -1)d)  (2a  (n -1)d)  ...  (2a  (n -1)d)
On the RHS we now have n lots of (2a  (n -1)d).
Therefore we have :
2Sn  n[2a  (n -1)d]
Dividing across be 2 we get :
Sn 
n
2
[2a  (n  1)d]
ARITHMETIC SEQUENCES
AND SERIES
Using the Sum Formula
Example:
Find the sum of the series
4 + 9 + 14 + 19 + … to 20 terms
Solution:
First term, a = 4
Common Difference, d = 5
Number of terms, n = 20
Substitute into formula: Sn 
S20 
20
2
2(4)  (20 -1)5
S20  10[8  19(5)]
S20  10[8  95]
S20  10[103]
S20  1030
n
2
2a  (n -1)d
ARITHMETIC SEQUENCES
AND SERIES
Using the Sum Formula
Example:
Find the sum of the series
2 + 6 + 10 + 14 + … + 158
Solution:
First term, a = 2
Common Difference, d = 4
Number of terms, n = ?
We need to find n before we can find the sum.
Un = 158
Un = a + (n -1)d
158 = 2 + (n – 1)(4)
158 = 2 + 4n – 4
158 = 4n – 2
160 = 4n
40 = n
cont’d on next slide
ARITHMETIC SEQUENCES
AND SERIES
Using the Sum Formula
Example Continued:
Find the sum of the series
2 + 6 + 10 + 14 + … + 158
Solution Continued:
First term, a = 2
Common Difference, d = 4
Number of terms, n = 40
We can now substitute into the sum formula to
give:
S40 
40
2
2(2)  (40 -1)4
S40  20[4  39(4)]
S40  20[4  156]
S40  20[160]
S40  3200
ARITHMETIC SEQUENCES
AND SERIES
Real-Life Problems
Arithmetic Series are used in everyday life to
solve problems.
Example:
Ahmed plans to save £250 in the year 2001,
£300 in 2002, £350 in 2003, and so on until the
year 2020.
Is savings for and arithmetic series with common
difference £50.
a.
Find the amount he plans to save in 2011
b.
Calculate his total planned savings over the
20-year period from 2001 to 2020.
Solution on next slide.
ARITHMETIC SEQUENCES
AND SERIES
Real-Life Problems
Solution:
a. a = £250
d = £50
From 2001 to 2011 is 11 years which means we
want the 11th term in the series.
2011: £250 + (11 – 1)(£50)
= £250 + 10(£50)
= £250 + £500
= £750
b. a = £250
S20 
20
2
d = £50
2(£250)  (20 -1)£50
S20  10[£500  19(£50)]
S20  10[£500  £950]
S20  10[£1450]
S20  £14 500
n = 20
ARITHMETIC SEQUENCES
AND SERIES
Assignment 4 – Sum of Arithmetic Series
Arithmetic Series in the Moodle Course Area.
This is a Yacapaca Activity.
Completed assignments must be submitted by
5:00pm on Monday 12th October.

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