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# 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 a1, a2, a3,...,an,... 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 an of such a sequence is given by the equation: an = a1 + (n - 1)d
Thus, a11 = a1 + (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 an of an arithmetic sequence is given by the equation:
an = a1 + (n - 1)d, in which d = an - an-1
a2 = a1 + (2 - 1)d = a1 + d = -2 ... (1)
a5 = a1 + (5 - 1)d = a1 + 4d = -17 ... (2)
(2) - (1) obtain 3d = -17 - (-2) = -17 + 2 = -15
d = -15/3 = -5 ... (3)
from (3) and (1) obtain a1 = -d - 2 = -(-5) - 2 = 5 - 2 = 3
Thus, a7 = a1 + (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 a1, a2, a3,..., an,... 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 a1, a1r, a1r2,...,a1rn-1,...for every positive integer n is a geometric sequence. The number r is called the common ratio, and the nth term an is given by an = a1rn-1
Since a1 = 3 and a2 = 9 , so the commom ratio r = a2/a1 = 3
an = a1rn-1 = 3 × 3n-1 = 31+n-1 = 3n and 729 = 36
Thus 3n = 36 , 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
Sn = (a1 + an)n/2
For the arithmetic sequence whose the nth term is
an = a1 + (n - 1)d
a10 = a1 + (10 - 1)5 = -3 + 9 × 5 = -3 + 45 = 42
Sn = (a1 + an)n/2 = ( -3 + 42 ) × 10/2 = 39 × 10/2 = 195
Therefore, the sum of the first 10 terms is 195.