Arithmetic and Geometric Sequences

Arithmetic and geometric are particular sequences which are used to solve problems involving number sequences which have a special pattern. With arithmetic sequences, the pattern is that the same number, called the common difference, is always added to the proceeding one, as in:
3, 7,11, 15, 19, . . .

With geometric sequences, the pattern is that the same number, called the common ratio, is always multiplied to the proceeding number, as in:

3, 6, 12, 24, 48, . . .

Because of these constant patterns, much is known about each type of sequence.

Arithmetic Sequences | Geometric Sequences | Sum Formulas | Example

Arithmetic Sequences

Any term of an arithmetic sequence with its common difference, denoted by d, can be predicted using an explicit formula, since
= + d

= + d = ( + d) + d = + 2 d

and so forth - each time adding one more "d" to obtain the next number.

So the explicit formula to generate any term of the sequence becomes:

= + ( n - 1 ) * d

Geometric Sequences

Any term of an geometric sequence with its common ratio, denoted by r, can be predicted using an explicit formula, since
= * r

= * r = ( * r ) * r = *

and so forth - each time multiplying one more "r" to obtain the next number.

So the explicit formula to generate any term of the sequence becomes:

= *

Sum Formulas

Both the arithmetic and geometric sequences have what are called sum formulas, which allow us to obtain the sum of the first n terms of either sequence.

The sum of the first n terms of an arithmetic sequence can be found by:

while the sum of the first n terms of a geometric sequence can be found by:

The derivations of these formulas are on pages 258 (arithmetic) and 259 (geometric) in the text.

Example

Thomas Malthus in 1798 said: "that population, when unchecked, increased in a geometric ration, and subsistence for man in an arithmetical ratio." How does this work out? Suppose we assume that:
1) the population is growing at a rate of 2% per year
2) it takes 1 acre to provide food for one person
3) the world has 8 billion acres of arable land
4) the world population is 1975 was 4 billion,

How long would it be before the world reaches maximum population?

Since the population is growing at the rate of 2%, the common ratio would be 1.02. The question then becomes for what value of n would:

OR

Using a calculator, approximately, when n = 35. So adding the 35 to 1975 = 2010 which is when the world would reach maximum population which would also mean massive starvation because of distribution problems.

 

 Week 4