Tutorial on sequences and summations.
A  Arithmetic Sequences
An arithmetic sequence is a sequence of numbers that is obtained by adding a constant number to the preceding number. The constant number is called the common difference.
Example 1:
0,6, 12, 18, 24, ... each term of the sequence is obtained by adding 6 to the preceding term.
If a_{1} is the first term and d is the common difference, the n_{th} term an is given by
a_{n} = a_{1} + (n  1)d
The sum S_{n} of the first n terms of an arithmetic sequence is given by
S_{n} = (n/2)(a_{1} + a_{n}) = (n/2)[2 a_{1} + (n  1)d]
Examples with Detailed Solutions
Example 2:
Which term of the arithmetic sequence 2, 5, 8... is equal to 227?
Solution to Example 2:
The first term a_{1} is 2 and the common difference is equal to: 5  2 = 8  5 = 3
Hence using the formula for the nth term, a_{n} = a_{1} + (n  1)d to the term equal to 227, we can write the equation:
227 = 2 + (n  1)3
Solve the above for n
n  1 = (227  2) / 3 = 75 and n = 76
The 76th term is equal to 227.
Example 3:
How many consecutive odd integers of an arithmetic sequence, starting from 9, must be added in order to obtain a sum of 15,860?
Solution to Example 3:
The first term a_{1} = 9 and d = 2 (the difference between any two consecutive odd integers). Hence the sum S_{n} of the n terms may be written as follows
S_{n} = (n/2)[2*a_{1} + (n  1)d] = 15,960
With a_{1} = 9 and d=2, the above equation in n may be written as follows
n^{2} + 8 n  15860 = 0
Solve the above for n
n = 122 and n = 130
The solution to the problem is that 122 consecutive odd numbers must be added in order to obtain a sum of 15,860.
Example 4:
What are the first 3 terms of an arithmetic sequence whose sum of the first n terms is equal to 2 n^{2} + 5 n?
Solution to Example 4:
The sum S_{n} of the first n terms is given by S_{n} = 2 n^{2} + 5 n. The sum of the first n1 terms is given by S_{n1} = 2(n1)^{2} + 5(n1). The difference between S_{n} and S_{n1} gives the nth term. Hence
a_{n} = S_{n}  S_{n1} = 2 n^{2} + 5 n  2(n1)^{2}  5(n1) = 2 n^{2} + 5 n  2 n^{2} + 4 n  2  5 n + 5
Which after simplification gives a_{n} = = 4 n + 3
We now use the formula for a_{n} to find the first 3 terms a_{1}, a_{2} and a_{3}.
a_{1} = 4(1)+3 = 7 , a_{2}= 4(2)+3 = 11 and a_{3} = 4(3)+3 = 15.
Example 5:
The sum of three consecutive numbers in an arithmetic sequence is equal 27 and their product is equal to 585. What are the three numbers?
Solution to Example 5:
Let the three numbers be x, x+d and x+3d where d is the common difference. Their sum is 27. Hence
x + (x+d) + (x+2d) = 27
Which gives 3x + 3d = 27 or x + d = 9
Their product is equal to 585. Hence
x(x+d)(x+2d) = 585
The equation x + d = 9 may be written as x = 9  d
Substitute x by 9  d in the equation x(x+d)(x+2d) = 585 to obtain
(9d)(9)(9+d) = 585
Simplify to obtain
9^{2}  d^{2} = 65
which gives d^{2} = 16
Solve to obtain d = 4 or d = 4
Use x = 9  d to obtain x = 5 for d = 4 and x = 13 for d = 4
There are two solutions to the given problem
1) d = 4 and the three terms are: 5, 9 and 13
2) d = 4 and the three terms are: 13, 9 and 5
Example 6:
The first three terms of an arithmetic sequence are as follows: x , 5 x/4 , 9/2. Find the first term x and the common difference d.
Solution to Example 6:
The common difference d is given by:
d = 5 x/4  x = 9/2  5 x/4
Which gives 5 x/4  x = 9/2  5 x/4
Solve the above equation for x
Common denominator: 5x/4  4x/4 = 18/2  5x/4
The equation: 6x = 18 and x = 3
d = 5 x / 4  x = 15/4  3 = 3/4
