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 a_{1} is the first term and r is the common difference, the n_{th} term an is given by
a_{n} = a_{1} × r^{n1}
The sum S_{n} of the first n terms of an arithmetic sequence is given by
S_{n} = a_{1}(1  r^{n}) / (1  r)
If the common ration r is such that r < 1 and the sum is infinite, then the sum is given by
S_{∞} = a_{1} / (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} = a_{1} r^{ 41} = a_{1} r^{ 3} = 1/8
a_{7} = a_{1} r^{ 71} = a_{1} r^{ 6} = 1/64
Then: a_{7} / a_{4} = a_{1} r^{ 6} / a_{1} r^{ 3} = r^{ 3} = (1/64) / (1/8) = 1/8
Hence r = 1/2
The 9th term: a_{9} = a_{1} r^{ 91} = a_{1} r^{ 8} = a_{1} r^{ 6} × r^{ 2} = a_{7} (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} = a_{1}(1  r^{n}) / (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 x^{3} + 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 = ( x^{3} + x ) / ( x^{2} + 1 )
The cross product gives
(x^{2} + 1) (x^{2} + 1) = x ( x^{3} + 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 / x^{2} + 4
Cross product gives
(x^{2} + 4)^{2} = 16 x^{2}
Expand and simplify
x^{4} + 8x^{4} + 16 = 16 x^{2}
x^{4}  8x^{4} + 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
