ASSIGNMENT 4: SUM OF ARITHMETIC SERIES
Question 1: The series, 2 + 5 + 8 + 11 + ... sums to 155. How many terms are
there in the series?
Solution:
a=2
d=3
Sn = 155
Substituting into the formula Sn = n/2[2a + (n-1)d] we get:
n/2[2(2) + (n-1)(3)] = 155
Multiply across by 2 to get:
n[4 + 3n – 3] = 310
n[1 + 3n] = 310
n + 3n2 = 310
3n2 + n - 310 =0
We now have a quadratic equation to solve either by factorisation or using the
formula.
Using the formula we get:
n
 1  12  4(3)(310)
2(3)
 1  1  3720
6
 1  61
n
6
n  10 or n  62/6
n
As the number of terms in a series can only be positive we take n = 10.
Question 2: The sum of the 55 terms in an arithmetic series is 7645 and the
last term is 274. Find the first term in the series.
Solution:
a=?
d=?
n = 55
Sn = 7645
55th term = 274
As we have two unknown variables, a and d, we will need to form two equations
and solve simultaneously.
Taking the information for the sum of the series and substituting into the sum
formula we get:
55/2[2a  (55  1)d]  7645
55/2[2a  54d]  7645
Multiply across by 2 to get:
55[2a  54d]  15290
110a  2970d  15290
2a  54d  278
Equation 1: 2a + 54d = 278
Taking that the 55th term is 274 we get:
a + (55-1)d = 274
Equation 2: a + 54d = 274
Solving equation 1 and 2 simultaneously:
Take (2) away from (1) to get
a = 4
Question 3: The first term of an arithmetic series is -10 and the common
difference is 8. What is the sum of the first 12 terms of the series?
Solution:
a = -10
d=8
n = 12
To find the sum of the first 12 terms we substitute the information we have
into the sum formula to get:
Sn  12/2[2(-10)  (12  1)(8)]
Sn  6[-20  88]
Sn  6[68]
Sn  408
Question 4: An arithmetic series has a = 60, d = -3 and n = 25. What is the sum
of the series?
Solution:
a = 60
d = -3
n = 25
To find the sum of the 25 terms in the series we substitute the values into the
sum formula to get:
Sn  25/2[2(60)  (25  1)(-3)]
Sn  25/2[120 - 72]
Sn  25/2[48]
Sn  600
Question 5: The first term of a series is 2 and the common difference is 4.
What is the sum of the first 23 terms?
Solution:
a=2
d=4
n = 23
To find the sum of the 23 terms in the series we substitute the values into the
sum formula to get:
Sn  23/2[2(2)  (23  1)(4)]
Sn  23/2[4  88]
Sn  23/2[92]
Sn  1058
Question 6: Find the sum of the series, 2 + 5 + 8 + 11 + ... + 62.
Solution:
a=2
d=3
n=?
Sn = ?
We have two unknowns in this question, the sum and the number of terms. To
work out the sum of the series we need to know the number of terms. As we
know a and d and the value of the last term we can work out n:
62 = 2 + (n-1)(3)
60 = 3n – 3
63 = 3n
21 = n
We can now find the sum of the series:
Sn  21/2[2(2)  (21  1)(3)]
Sn  21/2[4  60]
Sn  21/2[64]
Sn  672