back to
Algebra
Sequence Examples
Sequences Application 1
Question:
Find the eleventh term of the arthmetic sequences whose first fifth terms are -2, 5, 12, 19 and 26
Solution
A sequence a
1
, a
2
, a
3
,...,a
n
,... is called an arithmetic sequence if each term differs from the preceding term by a fixed amount.
The common difference of the given sequence is d = 7. The nth term a
n
of such a sequence is given by the equation: a
n
= a
1
+ (n - 1)d
Thus, a
11
= a
1
+ (n - 1)d = -2 + (11 - 1)7 = -2 + 10 × 7 = -2 + 70 = 68
Therefore, the eleventh term of this sequence is 68.
Sequences Application 2
Question:
If the second term of an arithmetic sequence is -2 and the fifth term is -17, find the seventh term.
Solution
The nth term a
n
of an arithmetic sequence is given by the equation:
a
n
= a
1
+ (n - 1)d, in which d = a
n
- a
n-1
a
2
= a
1
+ (2 - 1)d = a
1
+ d = -2 ... (1)
a
5
= a
1
+ (5 - 1)d = a
1
+ 4d = -17 ... (2)
(2) - (1) obtain 3d = -17 - (-2) = -17 + 2 = -15
d = -15/3 = -5 ... (3)
from (3) and (1) obtain a
1
= -d - 2 = -(-5) - 2 = 5 - 2 = 3
Thus, a
7
= a
1
+ (7 - 1)d = 3 + 6 × (-5) = 3 + (-30) = -27
Therefore, the seventh term is -27.
Sequences Application 3
Question:
If the two terms of a geometric sequence are 3 and 9, respectively, which term of the sequence is equal to 729?
Solution
A sequence a
1
, a
2
, a
3
,..., a
n
,... is called a geometric sequence if each term is obtained by multiplying the preceding term by a fixed amount.
In general, a sequence of the form a
1
, a
1
r, a
1
r
2
,...,a
1
r
n-1
,...for every positive integer n is a geometric sequence. The number r is called the common ratio, and the nth term a
n
is given by a
n
= a
1
r
n-1
Since a
1
= 3 and a
2
= 9 , so the commom ratio r = a
2
/a
1
= 3
a
n
= a
1
r
n-1
= 3 × 3
n-1
= 3
1+n-1
= 3
n
and 729 = 3
6
Thus 3
n
= 3
6
, so n = 6
Therefore, the 6th term of the sequence is 729.
Sequences Application 4
Question:
Find the sum of the first 10 terms of arithmetic sequence whose first term is -3 and whose common difference is 5.
Solution
The formula for the sum of the first n terms of an arithmetic sequence is
S
n
= (a
1
+ a
n
)n/2
For the arithmetic sequence whose the nth term is
a
n
= a
1
+ (n - 1)d
a
10
= a
1
+ (10 - 1)5 = -3 + 9 × 5 = -3 + 45 = 42
S
n
= (a
1
+ a
n
)n/2 = ( -3 + 42 ) × 10/2 = 39 × 10/2 = 195
Therefore, the sum of the first 10 terms is 195.