Solving Arithmetic Sequences

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 Sparks Rees