SOLVING ARITHMETIC SEQUENCES
The following examples are problems involving arithmetic sequences. I included here several sample problems with their solutions.
Problem Number One :
If the first three terms of an arithmetic sequence are 2, 6 and 10, find the 40th term.
To solve the problem we use this formula for finding the nth term of an arithmetic sequence.
An = A + (n – 1) d
Where, An = is the nth term, in the case of our problem it is the 40th term
A = the first term of the sequence , in our problem it is 2.
n = number of terms, in our problem it is 40.
d = the interval of the terms, or the difference of the next term
from the previous term, To get d; d = 6 – 2 = 4.
Now, it is time to substitute the values to the formula for solving nth term where the 40th term is to be solved.
An = 2 + (40 – 1 ) 4
An = 2 + (39) 4
An = 2 + 156
An = 158.
The 40th term of the arithmetic sequence is 158.
Problem Number Two :
If the first term of an arithmetic sequence is -3 and the eighth term is 11, find d and write the first 10 terms of the sequence.
In this problem,
A = -3 n = 8 A8 = 11
If these values are substituted in the formula for An, we have
11 = -3 + (8 – 1) d
11 = -3 + 7d
14 = 7d
d = 2
The first ten terms are -3, -1, 1, 3, 5, 7, 9, 11, 13, 15
SUM OF AN ARITHMETIC SEQUENCE
The sum of the first n terms of an arithmetic sequence with first term A and nth term An is;
Sn = n/2 (A + An) or this formula maybe rewritten as
Sn = n {(A+An)/2}
It can be remembered easily in this form as : “the number of terms multiplied by the mean value or average of the first and last terms.”
For an arithmetic sequence with the first term A and common difference d, the sum of the first n terms is ;
Sn = n/2 { 2a + (n – 1 ) d }
Problem Number Three :
Find the sum of all the odd integers from 1 to 1111, inclusive.
Solution :
Since the odd integers 1, 3, 5, etc, taken in order from the arithmetic sequence with d = 2, we can first find n from the formula for the nth term;
· 1111 = 1 + (n – 1) 2
1111 = 2n -1
1112 = 2n
n = 556
S = 556/2 ( 1 + 1111)
= 278 ( 1112)
= 309, 136
Problem Number Four :
If A = 4, n = 10, A10 = 49; find d and Sn.
Substituting the given values for A, n, and An in the formula:
An = A + (n – 1) d, we get
49 = 4 + (10 – 1 ) d
49 = 4 + 9d
45 = 9d
d = 5
By using Sn = n {( A + An)/2}, we have
S10 = 10 { (4 + 49 )/2} = 5 * 53 = 265
Source :
COLLEGE ALGEBRA (tenth edition ) by :
Paul K. Rees
Fred W. Sparks
Charles
