A
Number and Algebra
6
Arithmetic and geometric progressions
6.1
Arithmetic progressions
When a sequence has a constant difference between
successive terms it is called an
arithmetic progres-
sion
(often abbreviated to AP).
Examples include:
(i) 1, 4, 7, 10, 13,
. . .
where the
common differ-
ence
is 3 and
(ii)
a
,
a
+
d
,
a
+
2
d
,
a
+
3
d
,
. . .
where the common
difference is
d
.
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.
The sum
S
of an AP can be obtained by multi-
plying 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
[2
a
+
(
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.
The
n
’th term of an AP is
a
+
(
n
−
1)
d
The 6th term is:
a
+
5
d
=
17
(1)
The 13th term is:
a
+
12
d
=
38
(2)
Equation (2)
−
equation (1) gives: 7
d
=
21, from
which,
d
=
21
7
=
3.
Substituting in equation (1) gives:
a
+
15
=
17, from
which,
a
=
2.
Hence the 19th term is:
a
+
(
n
−
1)
d
=
2
+
(19
−
1)3
=
2
+
(18)(3)
=
2
+
54
=
56
.
Problem 3.
Determine the number of the term
whose value is 22 in the series 2
1
2
, 4, 5
1
2
, 7,
. . .
2
1
2
, 4, 5
1
2
, 7,
. . .
is
an
AP
where
a
=
2
1
2
and
d
=
1
1
2
.
52
NUMBER AND ALGEBRA
Hence if the
n
’th term is 22 then:
a
+
(
n
−
1)
d
=
22
i.e.
2
1
2
+
(
n
−
1)
(
1
1
2
)
=
22
(
n
−
1)
(
1
1
2
)
=
22
−
2
1
2
=
19
1
2
.
n
−
1
=
19
1
2
1
1
2
=
13 and
n
=
13
+
1
=
14
i.e.
the 14th term of the AP is 22
.
Problem 4.
Find the sum of the first 12 terms
of the series 5, 9, 13, 17,
. . .
5, 9, 13, 17,
. . .
is an AP where
a
=
5 and
d
=
4.
The sum of
n
terms of an AP,
S
n
=
n
2
[2
a
+
(
n
−
1)
d
]
Hence the sum of the first 12 terms,
S
12
=
12
2
[2(5)
+
(12
−
1)4]
=
6[10
+
44]
=
6(54)
=
324
Problem 5.
Find the sum of the first 21 terms
of the series 3.5, 4.1, 4.7, 5.3,
. . .
3.5, 4.1, 4.7, 5
.
3,
. . .
is an AP where
a
=
3
.
5 and
d
=
0
.
6.
The sum of the first 21 terms,
S
21
=
21
2
[2
a
+
(
n
−
1)
d
]
=
21
2
[2(3
.
5)
+
(21
−
1)0
.
6]
=
21
2
[7
+
12]
=
21
2
(19)
=
399
2
=
199.5
Now try the following exercise.
Exercise 28
Further problems on arith-
metic progressions
1. Find the 11th term of the series 8, 14, 20,
26,
. . .
[68]
2. Find the 17th term of the series 11, 10.7, 10.4,
10.1,
. . .
[6.2]
3. The seventh term of a series is 29 and the
eleventh term is 54. Determine the sixteenth
term.
[85.25]
4. Find the 15th term of an arithmetic progres-
sion of which the first term is 2.5 and the tenth
term is 16.
[23.5]
5. Determine the number of the term which is
29 in the series 7, 9.2, 11.4, 13
.
6,
. . .