Arithmetic progressions

Arithmetic progressions

Number sequences that have a constant difference between successive terms are called arithmetic progressions (AP).

For example:

  • 1, 3, 5, 7, 9, … is an arithmetic progression with a common difference of 2.
  • a, a+d, a+2d, a+3d, … is an arithmetic progression with a common difference of d.
  • 10, 20, 30, 40, … is an arithmetic progression with a common difference of 10.

In general the first term of an AP is represented by ‘a‘ and the common difference by ‘d‘ and the formulae for the nth term of the AP is:


nth term = a+(n − 1)d



Question 1

Find the 23rd term of the series 1, 5, 9, 13, …..

Solution 1

First we calculate the common difference, “d” by subtracting any two adjacent terms in the series.

d = 5-1
= 4

The first term, “a” is 1.

nth term = a+(n−1)d

23rd term = 1+(23-1)(4)
= 1+(22)(4)
= 1+88
= 89


Last term of an AP

Given the number of terms of a series, we can find the last term of the series using the formula nth term = a+(n−1)d and using using the number of terms in the series as n. Representing the last term by l, this becomes:


l = a+(n-1)d
for n terms.


Question 2

Find the last term of the series 1, 5, 9, 13,… given that the series contains 100 terms.

Solution 2

First we calculate the common difference, “d” by subtracting any two adjacent terms in the series.

d = 5-1
= 4

The first term, “a” is 1.

last term = a+(n−1)d

last term = 1+(100-1)(4)
= 1+(99)(4)
= 1+396
= 397


Sum of the terms of an AP

To find the sum S of an AP we multiply the average of all the terms by the number of terms.

average\ = \frac{first\ term + last\ term}{2}

Which means that average\ = \frac{a + l}{2}

Sum\ of\ terms = average \times number\ of\ terms

S_n = \frac{(a+l)n}{2}

S_n = \frac{n}{2}(a+l)

Substituting l = a+(n−1)d

We get S_n = \frac{n}{2}(a+a+(n−1)d)

Therefore sum of an AP of n terms:


S_n = \frac{n}{2}(2a+(n−1)d)


Question 3

Find the sum of the series 1, 5, 9, 13,… given that the series contains 100 terms.

Solution 3

S_n = \frac{n}{2}(2a+(n−1)d)

S_{100} = \frac{100}{2}[2(1)+(100−1)(4)]

S_{100} = 50[2+99(4)]

S_{100} = 50[2+396]

S_{100} = 50[398]

S_{100} = 19900


Worked Examples


Question 4

Determine (a) the 3rd, and (b) the 23rd term of the series 3, 8, 13, 18,…

Solution 4

a) d = 8-3
= 5

a = 3

nth term = a+(n−1)d

3rd term = 3+(3-1)(5)
= 1+(4)(5)
= 1+20
= 21

b) d = 8-3
= 5

a = 3

nth term = a+(n−1)d

23rd term = 3+(23-1)(5)
= 1+(22)(5)
= 1+110
= 11


Question 5

The 3rd term of an AP is 23 and the 23rd term is 87. Determine the 20th term.

Solution 5

nth term = a+(n−1)d

Using the 3rd term: 23 = a+(3−1)d
Using the 23rd term: 87 = a+(23-1)d

23 = a+2d ——(equation 1)
87 = a+22d —–(equation 2)

(eqn 2) minus (eqn 1)

64 = 20d

d = 3.2

Using equation 1: 23 = a+2d
23 = a+2(3.2)
23 = a+6.4
23-6.4 = a
a = 16.6

Hence 19th term = 16.6+(19−1)(3.2)
= 16.6+(18)(3.2)
= 16.6+57.6
= 74.2


Question 6

Determine the position of the term whose value is 22 in the series 2.5, 4, 5.5, 7…

Solution 6

a = 2.5
d = 1.5
and the nth term is 22 which means:

22 = a+(n−1)d
22 = 2.5+(n-1)(1.5)
22 = 2.5+1.5n-1.5
22 = 1+1.5n
22-1 = 1.5n
21 = 1.5n

n = 14

Therefore 22 is the 14th term.


Question 7

Find the sum of the first 42 terms of the series 1, 11, 21, 31,…

Solution 7

a = 1
d = 10
n = 42

S_n = \frac{n}{2}(2a+(n−1)d)

S_{42} = \frac{42}{2}[2(1)+(42−1)(10)]

S_{42} = 21[2+(41)(10)]

S_{42} = 21(2+410)

S_{42} = 21(412)

S_{42} = 8652


Question 8

Find the 11th term of the series 8, 14, 20, 26,… .

Solution 8

a = 8
d = 14-8 = 6

nth term = a+(n-1)d

11th term = 8+(11-1)(6)

= 8+10(6)

= 8+60

= 68


Question 9

Find the 17th term of the series 11, 10.7, 10.4, 10.1,… .

Solution 9

a = 11
d = 10.7-11 = -0.3

nth term = a+(n-1)d

17th term = 11+(17-1)(-0.3)

= 11+(16)(-0.3)

= 11-4.8

= 6.2


Question 10

The seventh term of a series is 29 and the eleventh term is 54. Determine the sixteenth term.

Solution 10

nth term = a+(n-1)d

7th term => 29 = a+(7-1)d

11th term => 54 = a+(11-1)d

29 = a+6d and 54 = a+10d

Then solve the simultaneous equations to get:

d = 6.25 and a = -8.5

Therefore 16th term = -8.5+(16-1)(6.25)

= -8.5+(15)(6.25)

= 85.25


Question 11

Find the 15th term of an arithmetic progression of which the first term is 2.5 and the tenth term is 16.

Solution 11

a = 2.5
10th term = 16

nth term = a+(n-1)d

10th term => 16 = 2.5+(10-1)d

16 = 2.5+9d

13.5 = 9d

d = 1.5

Therefore 15th term = 2.5+(15-1)(1.5)

= 2.5+14(1.5)

= 2.5+21

= 23.5


Question 12

Determine the number of the term which is 29 in the series 7, 9.2, 11.4, 13.6,… .

Solution 12

a = 7
d = 2.2
nth term = 29

nth term = a+(n-1)d

29 = 7+(n-1)(2.2)

29 = 7+2.2n-2.2

29 = 4.8+2.2n

29-4.8 = 2.2n

24.2 = 2.2n

n = 11


Question 13

Find the sum of the first 11 terms of the series 4, 7, 10, 13,… .

Solution 13

a = 4
d = 3
n = 11

S_n = \frac{n}{2}[2a+(n−1)d]

S_{11} = \frac{11}{2}[8+(11−1)(3)]

S_{11} = 5.5[8+(10)(3)]

S_{11} = 5.5(8+30)

S_{11} = 5.5(38)

S_{11} = 209


Question 14

Determine the sum of the series 6.5, 8.0, 9.5, 11.0, … , 32.

Solution 14

a = 6.5
d = 1.5
l = 32

l = a+(n-1)d

32 = 6.5+(n-1)(1.5)

32 = 6.5+1.5n-1.5

32 = 5+1.5n

32-5 = 1.5n

1.5n = 27

n = 18

S_n = \frac{(a+l)n}{2}

S_{18} = \frac{(6.5+32)(18)}{2}

S_{18} = \frac{(38.5)(18)}{2}

S_{18} = 346.5