52 Higher Engineering Mathematics
The n’th term of an AP is a + (n −1)d
The 6th term is: a + 5d = 17
(1)
The 13th term is: a + 12d= 38
(2)
Equation (2) −equation (1) gives: 7d = 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 .
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
[2a + (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
[2a + (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 24 Further problems on arithmetic
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 progression
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, ...
[11 th ]
6. Find the sum of the first 11 terms of the series
4, 7, 10, 13, ...
[209]
7. Determine the sum of the series 6.5, 8.0, 9.5,
11.0, ..., 32
[346.5]
6.3 Further worked problems on
arithmetic progressions
Problem 6. The sum of 7 terms of an AP is 35
and the common difference is 1.2. Determine the
first term of the series.
n = 7, d = 1.2 and S 7 = 35
Since the sum of n terms of an AP is given by
S n =
n
2
[2a + (n − 1)d], then
35 =
7
2
[2a + (7 − 1)1.2] =
7
2
[2a + 7.2]
The n’th term of an AP is a + (n −1)d
The 6th term is: a + 5d = 17
(1)
The 13th term is: a + 12d= 38
(2)
Equation (2) −equation (1) gives: 7d = 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 .
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
[2a + (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
[2a + (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 24 Further problems on arithmetic
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 progression
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, ...
[11 th ]
6. Find the sum of the first 11 terms of the series
4, 7, 10, 13, ...
[209]
7. Determine the sum of the series 6.5, 8.0, 9.5,
11.0, ..., 32
[346.5]
6.3 Further worked problems on
arithmetic progressions
Problem 6. The sum of 7 terms of an AP is 35
and the common difference is 1.2. Determine the
first term of the series.
n = 7, d = 1.2 and S 7 = 35
Since the sum of n terms of an AP is given by
S n =
n
2
[2a + (n − 1)d], then
35 =
7
2
[2a + (7 − 1)1.2] =
7
2
[2a + 7.2]