Chapter 6
Arithmetic and geometric
progressions
6.1 Arithmetic progressions
When a sequence has a constant difference between
successive terms it is called an arithmetic progression
(often abbreviated to AP).
Examples include:
(i) 1, 4, 7, 10, 13, ...where the common difference
is 3 and
(ii) a, a + d, a + 2d, a + 3d,...where the common
difference is d.
General expression for the n’th term of an AP
If the first term of an AP is ‘a’ and the common
difference is ‘d’ then
the n’th term is: a + (n − 1)d
In example (i) above, the 7th term is given by 1 +
(7 − 1)3 = 19, which may be readily checked.
Sum of n terms of an AP
The sum S of an AP can be obtained by multiplying the
average of all the terms by the number of terms.
The average of all the terms =
a +l
2
, where ‘a’ is the
first term and l is the last term, i.e. l = a + (n − 1)d, for
n terms.
Hence the sum of n terms,
S n = n
a + l
2
=
n
2
{a + [a + (n − 1)d]}
i.e.
S n =
n
2
[2a + (n − 1)d]
For example, the sum of the first 7 terms of the series 1,
4, 7, 10, 13, ... is given by
S 7 =
7
2
[2(1) + (7 − 1)3], since a = 1 and d = 3
=
7
2
[2 + 18] =
7
2
[20] = 70
6.2 Worked problems on arithmetic
progressions
Problem 1. Determine (a) the ninth, and (b) the
sixteenth term of the series 2, 7, 12, 17, ...
2, 7, 12, 17, ... is an arithmetic progression with a
common difference, d, of 5.
(a) The n’th term of an AP is given by a + (n −1)d
Since the first term a = 2, d = 5 and n =9 then the
9th term is:
2 + (9 −1)5 = 2 + (8)(5) = 2 + 40 =42
(b) The 16th term is:
2 + (16 −1)5 = 2 +(15)(5) = 2 + 75 =77.
Problem 2. The 6th term of an AP is 17 and the
13th term is 38. Determine the 19th term.