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Geometric progressions
You might have already heard the story about the wise man who invented chess and the reward he asked for from his sovereign. It was seemingly not much: he asked for \latex{ 1 } grain of rice for the first of the chess board's \latex{ 64 } squares, and he should receive twice as much for each subsequent square compared to the previous one. Therefore in total, he was given
\latex{s=1+2+2^{2}+2^{3}+2^{4}+...+2^{63}}
grains of rice for the whole board. An easy way to compute the total amount is by multiplying both sides by \latex{ 2 }:
\latex{2s=2+2^{2}+2^{3}+2^{4}+...+2^{64},}
and by subtracting the corresponding side of the original equation we get
\latex{s=2^{64}-1.}
The result is a huge number, written out in full it contains \latex{ 19 } digits.
The reward he each for each square is a term of a geometric sequence.
The reward he each for each square is a term of a geometric sequence.
DEFINITION: The sequence \latex{(a_{n} )} is called a geometric progression if there exists numbers a and q such that \latex{a_{1}=a} and \latex{a_{n+1}=a_{n}\times q.}
Number \latex{ q } is called the quotient or ratio of the geometric progression. The reason behind this name is that if \latex{a_{n}\neq 0} then equation \latex{a_{n+1}=a_{n}\times q} can also be written as follows:
\latex{\frac{a_{n+1} }{a_{n} }=q,}
meaning we can characterise a geometric progression by saying the quotient of any consecutive terms is constant.
If \latex{a_{1}=0} then all terms of the sequence are \latex{ 0 }, and \latex{ q } is arbitrary. Sometimes this sequence is not considered a geometric progression.
If \latex{a_{1}=0} then all terms of the sequence are \latex{ 0 }, and \latex{ q } is arbitrary. Sometimes this sequence is not considered a geometric progression.
If \latex{q=0} then \latex{a_{n}=0} for \latex{n\gt 1.}There is an alternative definition that also excludes this as a geometric progression. From now on, unless otherwise stated, we assume \latex{a_{1}\neq 0} and \latex{q\neq 0}
In the geometric progression presented in the introduction, \latex{a_{1}=1} and \latex{q=2.}
Example 1
Provide a formula for the \latex{n^{th}} term of the geometric progression \latex{(a_{n} )} using \latex{a_{1}}, \latex{ n } and \latex{ q }.
Solution
By computing an for some specific values of \latex{ n } \latex{(e.g.n=1,2,3,4)} we can state the following conjecture:
\latex{a_{n}=a_{1}\times q^{n-1}.}
The correctness of the formula can easily be verified by induction.
It is true for \latex{n=1:a_{1}=a_{1}.}
Suppose that it is true for \latex{ n }. We have to show that it is true for \latex{n+1} as well:
It is true for \latex{n=1:a_{1}=a_{1}.}
Suppose that it is true for \latex{ n }. We have to show that it is true for \latex{n+1} as well:
\latex{a_{n+1}=a_{n}\times q=a_{1}\times q^{n-1}\times q=a_{1}\times q^{n}.}
Therefore the conjecture is verified.
Terms of a geometric progression can be easily represented by the graph of the function \latex{f(x)=q\times x.} On Figures 16/a and 16/b the cases \latex{q=2,a_{1}=1} and \latex{q=\frac{1}{2},a_{1}} respectively are represented.
it is worth taking a look at the cases \latex{a_{1}=1,q=-2} (Figure 16/c) and \latex{a_{1}=1,q=-\frac{1}{2}} (Figure 16/d) as well.
Terms of a geometric progression can be easily represented by the graph of the function \latex{f(x)=q\times x.} On Figures 16/a and 16/b the cases \latex{q=2,a_{1}=1} and \latex{q=\frac{1}{2},a_{1}} respectively are represented.
it is worth taking a look at the cases \latex{a_{1}=1,q=-2} (Figure 16/c) and \latex{a_{1}=1,q=-\frac{1}{2}} (Figure 16/d) as well.
\latex{y=2x}
\latex{y=x}
\latex{y=x}
\latex{y=\frac{1}{2}\times x}
\latex{y=-2x}
\latex{y=x}
\latex{y=x}
\latex{y=-\frac{1}{2}\times x}
\latex{ a_4 }
\latex{ a_3 }
\latex{ a_2 }
\latex{ a_1 }
\latex{ 0 }
\latex{ 1 }
\latex{ x }
\latex{ 0 }
\latex{ 1 }
\latex{ x }
\latex{ a_3 }
\latex{ a_2 }
\latex{ a_1 }
\latex{ y }
\latex{ y }
\latex{ y }
\latex{ y }
\latex{ a_3 }
\latex{ a_1 }
\latex{ a_2 }
\latex{ a_4 }
\latex{ 1 }
\latex{ x }
\latex{ x }
\latex{ 1 }
\latex{ a_2 }
\latex{ a_1 }
\latex{ d })
\latex{ c })
\latex{ b })
\latex{ a })
Figure 16
Example 2
For which values of \latex{q\neq 0} is it true that the geometric progression \latex{(a_{n} )} satisfies the equation \latex{a_{n+2}=a_{n}+a_{n+1}} for \latex{(a_{1}\neq 0 )n\geq 1?}
Solution
Substitute the expression for the terms of a geometric progression into the equation:
\latex{a_{1}\times q^{n+1}=a_{1}\times q^{n-1}+a_{1}\times q^{n}.}
Since we have \latex{a_{1}\neq 0} and \latex{q\neq 0} we can divide through by \latex{a_{1}} and \latex{q^{n-1}:}
\latex{q^{2}=1+q}, and this, \latex{q^{2}-q-1=0}
The resulting quadratic equation yields
\latex{q_{1}=\frac{1+\sqrt{5} }{2} } and \latex{q_{2}=\frac{1-\sqrt{5} }{2}.}
By checking both values with an arbitrary \latex{a_{1}\neq 0,} we can see that it satisfies the equation described in the problem for any \latex{n\geq 1.}
It is worth noting that the equation described in the problem is the recursive formula defining the Fibonacci sequence.
It is worth noting that the equation described in the problem is the recursive formula defining the Fibonacci sequence.
Example 3
Provide a formula for the sum of the first \latex{ n } terms of the geometric progression \latex{(a_{n} )} , or in other words, find \latex{(s_{n} )} if
\latex{(s_{n} )=a_{1}+a_{2}+a_{3}+...+a_{n}.} (1)
Solution
The following idea might help. Multiply both sides of the equation by \latex{ q }.
\latex{q\times s_{n}=a_{1}\times q+a_{2}\times q+a_{3}\times q+...+a_{n}\times q=a_{2}+a_{3}+a_{4}+...+a_{n+1}} (2)
using the definition of the geometric progression. Now subtracting the relevant sides of equation (1) from those of equation (2) yields
\latex{q\times s_{n}=a_{1}\times q+a_{2}\times q+a_{3}\times q+...+a_{n}\times q=a_{2}+a_{3}+a_{4}+...+a_{n+1}} (2)
using the definition of the geometric progression. Now subtracting the relevant sides of equation (1) from those of equation (2) yields
\latex{s_{n}\times (q-1)=a_{n+1}-a_{1}=a_{1}\times (q^{n}-1 )}.
\latex{y=\frac{1}{2}\times x+1}
\latex{y=x}
\latex{a=1,q=\frac{1}{2}}
\latex{ 1 }
\latex{ x }
\latex{ 0 }
\latex{ 1 }
\latex{ 2 }
\latex{ y }
Figure 17
If \latex{q\neq 1,} then we have
\latex{s_{n}=a_{1}\times \frac{q^{n}-1 }{q-1} (q\neq 1).}
If \latex{q=1}, then obviously \latex{a_{1}=a_{2}=...=a_{n}=a,} and therefore equation (1) is:
\latex{s_{n}=n\times a.}
Note that \latex{s_{n}} can be written using a recursive definition:
\latex{s_{1}=a} and \latex{s_{n+1}=q\times s_{n}+a.}
By the latter definition, terms of the sequence \latex{(s_{n} )} can be represented using the graph of function \latex{f(x)=qx+a.} (Figure 17).
Example 4
We divide the unit square into \latex{ 4 } congruent squares, then we colour \latex{ 3 } squares red, blue and green, respectively (Figure 18). Now we divide the fourth square into \latex{ 4 } congruent squares, from which we colour \latex{ 3 } squares red, blue and green as seen in Figure 19. This process is repeated \latex{ n } times. What will be the total area of the red part?
Figure 18
Solution
In the first step one quarter, \latex{\frac{1}{4}} unit had been painted red. In each step, the area remaining white is as large as the red one, and in the next step \latex{\frac{1}{4}} th of it will be painted red. Therefore the total area painted red after \latex{ n } steps is:
\latex{\frac{1}{4}+\frac{1}{4^{2} }+\frac{1}{4^{3} }+...+\frac{1}{4}\times \frac{1-\frac{1}{4^{n} } }{1-\frac{1}{4} }=\frac{1}{3}\times \left(1-\frac{1}{4^{n} } \right).}
Figure 19
Example 5
Observe these two equations.
\latex{11-2=9=3^{2},} \latex{1,111-22=1,089=33^{2}.}
Is it true for an arbitrary integer \latex{n\gt 0} that \latex{111...1-222...2=33...3^{2},} where the number consisting of \latex{ 1s } has \latex{ 2n } digits, and the numbers consisting of \latex{ 2s } and \latex{ 3s } have \latex{ n } digits?
Solution
The number comprised of \latex{ 2n } \latex{ 1s } can be written as:
\latex{1+10+10^{2}+...+10^{2n-1}=\frac{10^{2n}-1}{9}.}
Similarly, the number comprised of \latex{ n } \latex{ 2s } can be written as:
\latex{2\times (1+10+10^{2}+...+10^{n-1} )=2\times \frac{10^{n}-1 }{9}=\frac{2\times 10^{n}-2 }{9}.}
Their difference is:
\latex{\frac{10^{2n}-1 }{9}-\frac{2\times 10^{n}-2 }{9}=\frac{10^{2n}-2\times 10^{n}+1 }{9}=\left(\frac{10^{n}-1 }{3} \right)^{2}.}
Furthermore the number comprised of \latex{ n } \latex{ 3s } can be written as:
\latex{3\times (1+10+10^{2}+...+10^{n-1} )=3\times \frac{10^{n}-1 }{9}=\frac{10^{n}-1 }{3}.}
Therefore the statement is true for any positive integer \latex{ n }.
Example 6
In a geometric progression, the sum of the first three terms is \latex{ 31 } and the sum of the first and the third term is \latex{ 26 }. Compute the first term and the quotient of the sequence.
Solution
We formulate the equations:
\latex{a_{1}+a_{2}+a_{3}=31;}
\latex{a_{1}+a_{3}=26.}
It follows that \latex{a_{2}=a_{1}\times q=5,} and from the second equation
\latex{a_{1}+a_{1}\times q^{2}=26.}
Since \latex{q\neq 0,a_{1}=\frac{5}{q},} substituting \latex{a_{1}} we get
\latex{\frac{5}{q}+5q=26,}
\latex{5q^{2}-26q+5=0.}
\latex{5q^{2}-26q+5=0.}
Therefore the possible values of \latex{ q } are \latex{ 5 } and \latex{\frac{1}{5},} while the corresponding \latex{ a1 } values are \latex{ 1 } and \latex{ 25 }, respectively.
Both solutions satisfy the criteria because in the first case the first three terms of the sequence are \latex{ 1;\, 5;\, 25; } and \latex{ 25;\, 5;\, 1 } in the second case.
Example 7
For four subsequent terms of a geometric progression, the sum of the two end terms is \latex{ 112 }, while that of the two in the middle is \latex{ 48 }. What is the quotient of the geometric progression?
Solution
Let \latex{ q } be the quotient of the geometric progression and \latex{ b } the first of the four terms. According the criteria:
\latex{b+b\times q^{3}=112} and \latex{b\times q+b\times q^{2}=48.}
By factorizing both sides of the equation we get
\latex{b\times (1+q)\times (1-q+q^{2} )=112} and \latex{b\times q\times (1+q)=48.}
Since \latex{q\neq 0} the second equation yields: \latex{b\times (1+q)=\frac{48}{q},} which can be substituted in the first equation to get
\latex{\frac{48}{q}\times (1-q+q^{2} )=112.}
By rearranging this gives us the following equation
\latex{3q^{2}-10q+3=0.}
The possible values of \latex{ q } are \latex{ 3 } and \latex{\frac{1}{3}}.
The values of q obtained are both solutions as the corresponding values of \latex{ b } are \latex{ 4 } and \latex{ 108 }, thus the two geometric progressions are
\latex{4;\,12;\,36;\,108;\,324...} and \latex{108;\,36;\,12;\,4;\frac{4}{3};\,...}
Example 8
Three numbers are consecutive terms of an increasing geometric sequence. Their sum is \latex{ 7 }. By subtracting \latex{ 1 } from the largest number, the resulting numbers are consecutive terms of an arithmetic progression. What are these numbers?
Solution
Let \latex{ a } be the smallest number and \latex{ q } the quotient of the geometric progression. Then the criteria can be written as:
\latex{a+aq+aq^{2}=7} and \latex{2aq=a+aq^{2}-1.}
From the second equation
\latex{a+aq^{2}=2aq+1,}
by substituting this into the first equation and rearranging we have
\latex{3aq=6,}
\latex{aq=2.}
\latex{aq=2.}
Again, substituting aq into the first equation and after rearranging,
\latex{a+2q=5,}
\latex{a=5-2q.}
\latex{a=5-2q.}
Now from the equation \latex{aq=2} we get
\latex{2q^{2}-5q+2=0.}
This means either \latex{q=2} or \latex{q=\frac{1}{2}.}
Since the sought geometric progression is increasing, only the solution \latex{q=2} is possible. Thus a equals \latex{ 1 } and the three numbers are: \latex{ 1;\, 2;\, 4 }.
Since \latex{ 1;\, 2;\, 3 } are indeed subsequent terms of an arithmetic progression, the result meets the criteria.
Since the sought geometric progression is increasing, only the solution \latex{q=2} is possible. Thus a equals \latex{ 1 } and the three numbers are: \latex{ 1;\, 2;\, 4 }.
Since \latex{ 1;\, 2;\, 3 } are indeed subsequent terms of an arithmetic progression, the result meets the criteria.
Example 9
Prove that if \latex{(a_{n} )} is a geometric progression, \latex{a_{n}\gt 0} and \latex{0\lt k\lt n} are integers, then \latex{a_{n}} is the geometric mean of numbers \latex{a_{n-k}} and \latex{a_{n+k}}, that is
\latex{a_{n^{2} }=a_{n-k}\times a_{n+k.}}
Solution
Use the known formula for the \latex{(n-k)^{th}} and the \latex{(n+k)^{th}} term, and then multiply the two terms.
\latex{a_{n-k}=a\times q^{n-k-1},}
\latex{a_{n+k}=a\times q^{n+k-1},}
\latex{a_{n-k}\times a_{n+k}=a^{2}\times q^{2n-2}=(a\times q^{n-1} )^{2}=a_{n^{2} }.}
\latex{a_{n+k}=a\times q^{n+k-1},}
\latex{a_{n-k}\times a_{n+k}=a^{2}\times q^{2n-2}=(a\times q^{n-1} )^{2}=a_{n^{2} }.}
Example 10
Three integers, whose sum is \latex{ 60 }, are consecutive terms of an arithmetic progression. By adding \latex{ 2.2,\, 4 } and \latex{ 7 } to these numbers respectively, the resulting numbers are consecutive terms of a geometric progression. What are the original integers?
Solution
Denote the number in the middle by \latex{ x }, and the difference of the arithmetic progression by \latex{ d }.
Then the three integers are: \latex{x-d,x,x+d}
Since their sum is \latex{ 60 },
Then the three integers are: \latex{x-d,x,x+d}
Since their sum is \latex{ 60 },
\latex{3x=60}
\latex{x=20}.
\latex{x=20}.
Thus the numbers are: \latex{20-d,20,20+d}
The three consecutive terms of the geometric progression are
\latex{22.2-d, 24 ,27+d.}
We know that the square of the number in the middle is the product of the two other terms:
\latex{(22.2-d)\times (27+d)=24^{2}.}
By expanding and rearranging we get the following equation:
\latex{d^{2}+4.8d-23.4=0.}
The two roots are: \latex{d_{1}=3,d_{2}=-7.8.} But since we are looking for integers, only \latex{d=3} is a solution, thus the three numbers are:
\latex{17;\,20;\,23}
These meet the criteria.
Exercises
{{exercise_number}}. In the geometric progression \latex{(a_{n} ) a_{5}=96, a_{6}=192.} Find the first term and the quotient.
{{exercise_number}}. Show that tangents of the numbers \latex{\frac{\Pi }{6},\frac{\Pi }{4},\frac{\Pi }{3}} are consecutive terms of a geometric progression.
{{exercise_number}}. The sum of the first three terms of a geometric progression is the \latex{\frac{1}{8}} th of the sum of the next three terms. What is the quotient of the sequence?
{{exercise_number}}. Find the sum of the first ten non-negative integer powers of \latex{ 2 }.
{{exercise_number}}. We know the following about the geometric progressions \latex{(a_{n} ):}
- \latex{a_{4}-a_{2}=18, \\ a_{5}-a_{3}=36;}
- \latex{a_{7}-a_{4}=-216, \\ a_{5}-a_{4}=-72;}
- \latex{a_{1}-a_{3}+a_{5} =-65, \\ a_{1}-a_{7}=-325.}
Compute the first term and the quotient of the sequences.
{{exercise_number}}. Prove that if a, b and c are three consecutive terms of a geometric progression, then
\latex{(a+b+c)\times (a-b+c)=a^{2}+b^{2}+c^{2}.}
{{exercise_number}}. Some numbers had been written in the following table. Fill in the empty slots such that the elements of each row and column are consecutive terms of a geometric progression.
\latex{ 27 }
\latex{ 36 }
\latex{ 6 }
\latex{ 8 }
{{exercise_number}}. The sum of the first three terms of a geometric progression is \latex{ 42 }. The same numbers are the first, second and sixth terms of an increasing arithmetic progression. What are these numbers?
{{exercise_number}}. The sum of the first three terms of an increasing arithmetic progression is \latex{ 54 }. If the first term is unchanged, the second one is reduced by \latex{ 9 } and the third by \latex{ 6 }, then we obtain three consecutive terms of a geometric progression. Determine the first term and the difference of the arithmetic progression.