Geometric Sequences and Sums

Tutorial on geometric sequences and summations.

A - Geometric Sequences

An arithmetic sequence is a sequence of numbers that is obtained by multiplying the preceding number by a constant number called the common ratio.

Example 1:

1,2, 4, 8, 16, ... each term of the sequence is obtained by multiplying by 2 the preceding term.

If a1 is the first term and r is the common difference, the nth term an is given by

an = a1 rn-1

The sum Sn of the first n terms of an arithmetic sequence is given by

Sn = a1(1 - rn) / (1 - r)

If the common ration r is such that |r| < 1 and the sum is infinite, then the sum is given by

S = a1 / (1 - r)



Examples with Detailed Solutions

Example 3:

The 4th and 7th terms of a geometric sequence are 1/8 and 1/64 respectively. Find the 9th term.

Solution to Example 2:

a
4 = a1 r 4-1 = a1 r 3 = 1/8

a
7 = a1 r 7-1 = a1 r 6 = 1/64

Then: a
7 / a4 = a1 r 6 / a1 r 3 = r 3 = (1/64) / (1/8) = 1/8

Hence r = 1/2

The 9th term: a
9 = a1 r 9-1 = a1 r 8 = a1 r 6 r 2 = a7 (1/2) 2 = (1/64)(1/4) = 1/256

Example 3:

Find 3 consecutive numbers in a geometric sequence whose sum is 234 and their product is 157,646.

Solution to Example 3:

Let the 3 terms be : x , x r and x r
2

sum = x + x r + x r
2 = x(1 + r + r 2) = 234

product = x(x r)(x r
2) = x 3 r 3 = 157464

The last equation gives: x r =
3 √ (157464) = 54 or x = 54/r

We now substitute x by 54/r in the equation x(1 + r + r
2) = 234 obtained above to get

(54/r)(1 + r + r
2) = 234 or 54(1 + r + r 2) = 234 r

Solve the above quadratic equation for r to obtain

r = 3 or r = 1/3

Two possible solutions to the given problem:

Use r = 3 and x r = 54 to find the three terms: x = 18, x r = 54 and x r
2 = 162

Use r = 1/3 and x r = 54 to find the three terms: x = 162, x r = 54 and x r
2 = 18
Example 4:

The first term of a geometric sequence is 2000 and the second term is 1000. Starting from the first term, how many consecutive terms in this sequence must be taken in order to obtain a sum equal to 3875?

Solution to Example 4:

First the common ration r = 1000/2000 = 1/2

In a geometric sequence the sum is given by

S
n = a1(1 - rn) / (1 - r) = 2000(1 - (1/2)n) / (1 - 1/2) = 3875

We rewrite the above equation as follows:

1 - (1/2)
n = ( 3875 (1/2) )/ 2000

which gives

(1/2)
n = 1 - ( 3875 (1/2) )/ 2000 = 125 / 4000 = 1/32 = (1/2)5

Which gives n = 5. The sum of the first five terms gives 3875


Example 5:

Prove that x, x
2 + 1 and x3 + x cannot be the 3 consecutive terms in a geometric sequence of real numbers.

Solution to Example 5:

Suppose they are the three terms are that of a geometric sequence and express the common ratio using the three terms and write the following equation

(x
2 + 1) / x = ( x3 + x ) / ( x2 + 1 )

The cross product gives

(x
2 + 1) (x2 + 1) = x ( x3 + x )

Expand and simplify

2 x
2 + 1 = 0

The above equation has no real solution and therefore the sequence as defined above cannot be a geometric sequence with real numbers.


Example 6:

The first three terms of an arithmetic sequence are as follows: x , x
2 + 4 and 16x. Find the three terms.

Solution to Example 6:

Use the three terms to find two expressions of the common ratio and equate them to obtain the equation

(x
2 + 4) / x = 16 x / x2 + 4

Cross product gives

(x
2 + 4)2 = 16 x2

Expand and simplify

x
4 + 8x4 + 16 = 16 x2

x
4 - 8x4 + 16 = 0

(x
2 - 4)2 = 0

x
2 - 4 = 0

x = 2 or x = -2

Two possible solutions:

1) x = 2, Hence the three terms are :x = 2, x
2 + 4 = 8 and 16 x = 32

2) x = -2, Hence the three terms are :x = -2, x
2 + 4 = 8 and 16 x = - 32

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Updated: 21 October 2014 (A Dendane)
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