Arithmetic and geometric progressions 53
Hence
35 × 2
7
= 2a + 7.2
10 = 2a + 7.2
Thus
2a = 10 − 7.2 = 2.8,
from which
a =
2.8
2
= 1.4
i.e. the first term, a = 1.4
Problem 7. Three numbers are in arithmetic
progression. Their sum is 15 and their product is 80.
Determine the three numbers.
Let the three numbers be (a − d), a and (a + d)
Then (a − d) + a + (a + d) = 15, i.e. 3a = 15, from
which, a = 5
Also, a(a − d)(a + d) = 80, i.e. a(a 2 − d 2 ) = 80
Since a = 5, 5(5
2
− d
2
) = 80
125 − 5d
2
= 80
125 − 80 = 5d
2
45 = 5d
2
from which, d 2 =
45
5
= 9. Hence d =
√
9 =±3.
The three numbers are thus (5 − 3), 5 and (5 + 3), i.e.
2, 5 and 8.
Problem 8. Find the sum of all the numbers
between 0 and 207 which are exactly divisible by 3.
The series 3, 6, 9, 12, ..., 207 is an AP whose first term
a = 3 and common difference d = 3
The last term is
a + (n − 1)d = 207
i.e.
3 + (n − 1)3 = 207,
from which
(n − 1) =
207 − 3
3
= 68
Hence
n = 68 + 1 = 69
The sum of all 69 terms is given by
S 69 =
n
2
[2a + (n − 1)d]
=
69
2
[2(3) + (69 − 1)3]
=
69
2
[6 + 204] =
69
2
(210) = 7245
Problem 9. The first, twelfth and last term of an
arithmetic progression are 4, 31
1
2 , and 376
1
2
respectively. Determine (a) the number of terms in
the series, (b) the sum of all the terms and (c) the
‘80’th term.
(a) Let the AP be a, a +d, a +2d, ... , a + (n − 1)d,
where a = 4
The 12th term is: a + (12 −1)d = 31
1
2
i.e.
4 + 11d = 31
1
2 ,
from which, 11d = 31
1
2 − 4 = 27
1
2
Hence d =
27
1
2
11
= 2
1
2
The last term is a + (n − 1)d
i.e. 4 + (n − 1)
2
1
2
= 376
1
2
(n − 1) =
376
1
2 − 4
2
1
2
=
372
1
2
2
1
2
= 149
Hence the number of terms in the series,
n = 149 +1 =150
(b) Sum of all the terms,
S 150 =
n
2
[2a + (n − 1)d]
=
150
2
2(4) + (150 − 1)
2
1
2
= 75
8 + (149)
2
1
2
= 85[8 + 372.5]
= 75(380.5) = 28537
1
2
(c) The 80th term is:
a + (n − 1)d = 4 + (80 − 1)
2
1
2
= 4 + (79)
2
1
2
= 4 + 197.5 = 201
1
2
Hence
35 × 2
7
= 2a + 7.2
10 = 2a + 7.2
Thus
2a = 10 − 7.2 = 2.8,
from which
a =
2.8
2
= 1.4
i.e. the first term, a = 1.4
Problem 7. Three numbers are in arithmetic
progression. Their sum is 15 and their product is 80.
Determine the three numbers.
Let the three numbers be (a − d), a and (a + d)
Then (a − d) + a + (a + d) = 15, i.e. 3a = 15, from
which, a = 5
Also, a(a − d)(a + d) = 80, i.e. a(a 2 − d 2 ) = 80
Since a = 5, 5(5
2
− d
2
) = 80
125 − 5d
2
= 80
125 − 80 = 5d
2
45 = 5d
2
from which, d 2 =
45
5
= 9. Hence d =
√
9 =±3.
The three numbers are thus (5 − 3), 5 and (5 + 3), i.e.
2, 5 and 8.
Problem 8. Find the sum of all the numbers
between 0 and 207 which are exactly divisible by 3.
The series 3, 6, 9, 12, ..., 207 is an AP whose first term
a = 3 and common difference d = 3
The last term is
a + (n − 1)d = 207
i.e.
3 + (n − 1)3 = 207,
from which
(n − 1) =
207 − 3
3
= 68
Hence
n = 68 + 1 = 69
The sum of all 69 terms is given by
S 69 =
n
2
[2a + (n − 1)d]
=
69
2
[2(3) + (69 − 1)3]
=
69
2
[6 + 204] =
69
2
(210) = 7245
Problem 9. The first, twelfth and last term of an
arithmetic progression are 4, 31
1
2 , and 376
1
2
respectively. Determine (a) the number of terms in
the series, (b) the sum of all the terms and (c) the
‘80’th term.
(a) Let the AP be a, a +d, a +2d, ... , a + (n − 1)d,
where a = 4
The 12th term is: a + (12 −1)d = 31
1
2
i.e.
4 + 11d = 31
1
2 ,
from which, 11d = 31
1
2 − 4 = 27
1
2
Hence d =
27
1
2
11
= 2
1
2
The last term is a + (n − 1)d
i.e. 4 + (n − 1)
2
1
2
= 376
1
2
(n − 1) =
376
1
2 − 4
2
1
2
=
372
1
2
2
1
2
= 149
Hence the number of terms in the series,
n = 149 +1 =150
(b) Sum of all the terms,
S 150 =
n
2
[2a + (n − 1)d]
=
150
2
2(4) + (150 − 1)
2
1
2
= 75
8 + (149)
2
1
2
= 85[8 + 372.5]
= 75(380.5) = 28537
1
2
(c) The 80th term is:
a + (n − 1)d = 4 + (80 − 1)
2
1
2
= 4 + (79)
2
1
2
= 4 + 197.5 = 201
1
2