We will now check out geometric sequences.
A geometric sequence is a sequence in which each term is obtained by multiplying the previous term with a fixed number.
This fixed number is called the common ratio. It is often denoted with #r#.
Hence, for every #n# greater than #1# we have: #t_n=r \cdot t_{n-1}#.
Hence, the terms in the sequence increase exponentially if #r>1#.
If #0<r<1# then the terms in the sequence decrease exponentially.
Give the initial term #t_1#, the common ratio #r# and the number of terms #n# of the following sequence #t#.
\[3, 12, 48, 192, \ldots, 49152, 196608\]
#t_1=3#
#r=4#
#n=9#
The initial term is the first term and is equal to #3#. Hence, #t_1=3#.
To calculate the common ratio, we divide to subsequent terms with each other: #\frac{12}{3}=4# (note: we divide the second term #t_2# by the first one #t_1# and not the other way around). We see that this common ratio also holds for other subsequent terms. Hence, it's a geometric series. In this case, we have #r=4#.
Since the sequence starts at #3# and ends at #196608#, in which each time is multiplied by #4#, this sequence has #9# terms. Hence, #n=9#.
Each sequence which is both arithmetic and geometric is constant.
This can be seen by using the first three terms of the sequence #t_1#, #t_2#, #t_3#:
- Since the sequence is arithmetic, there is a difference #d#, such that #t_2=t_1+d# and #t_3=t_1+2v#.
- Since the sequence is geometric, there is a common ratio #r# such that #t_2= t_1\cdot r# and #t_3=t_1\cdot r^2#.
If we compare the two expressions for #t_2# and next the two expressions for #t_3#, and after moving all terms with #t_1# to the left-hand side and all terms without #t_1# to the right-hand side, we find the system of equations:
\[\eqs{t_1\cdot(r-1) &=& d\\ t_1\cdot(r^2-1) &=& 2 d}\]
After multiplying both terms of the first equation by #(r+1)#, both equations are equal on the left-hand side. Hence, after equalizing the right-hand sides we get
\[(r+1)\cdot d = 2d\]
Hence, #d=0# or #r=1#. In both cases, the sequence is constant.
The formula in the definition for finding the #n#-th term is called a recursive formula since the previous term is needed. We can also compose a direct formula. That is a formula for finding the #n#-th term by means of the rank number and the common ratio. There we do not need the previous term.
The direct formula for calculating the #n#-th term of an geometric sequence #t_1, t_2, \ldots#, is: \[t_n=t_1\cdot r^{n-1}\] where #r# is the common ratio.
In general, the terms of a geometric sequence can be described as follows:
\[\begin{array}{lcl}
t_1&=&t_1 \\
t_2=t_1 \cdot r&=&t_1 \cdot r\\
t_3=t_2 \cdot r = t_1\cdot r \cdot r &=&t_1 \cdot r^2\\
t_4=t_3 \cdot r=t_1 \cdot r^2 \cdot r &=&t_1 \cdot r^3\\
\vdots \phantom{=t_2 \cdot r = t_1xx} &&\vdots\\
t_{20}=t_{19} \cdot r&=&t_1 \cdot r^{19}\\ \vdots \phantom{=t_2 \cdot r = t_1xx} &&\vdots
\end{array}\]
This leads us to the direct formula: \[t_n=t_1\cdot r^{n-1}\]
In this case, the sequence starts at #t_1#. We can also let the sequence start at #t_0#. In that case, the direct formula is: \[t_n=t_0\cdot r^n\]
This follows from the given formula since there are now #n+1# terms, and the initial term is #t_0#.
Notice that if we replace the addition by multiplication and multiplication by exponentiation in the direct formula for the term with rank number #n# of an arithmetic sequence, we end up with the direct formula for the term with rank number #n# of a geometric sequence.
In other words: the logarithm of the #n#-th term of a geometric sequence is the #n#-th term of an arithmetic sequence. In there, the difference of the arithmetic sequence is equal to the logarithm of the common ratio.
Here are a couple of examples which show how we can compose the direct formula and how it can be used to calculate a term.
Which direct formula corresponds with the following geometric sequence?\[a_1=2,\quad a_2=8{,}\quad a_3=32{,}\quad a_4=128{,}\quad\ldots\]
#a_k=# #2\cdot 4^{k-1}#
Given is the fact that #a_k# is a geometric sequence. This means that the direct formula is equal to #a_k=a_1 \cdot r^{k-1}#, in which #r# is the common ratio. hence, we have to determine #a_1# and the common ratio.
- The initial term #a_1# is given and equal to #2#.
- The common ratio is equal to #r=\frac{8}{2}=4#.
Consequently, the direct formula gives \[a_k=2\cdot 4^{k-1}\]
Geometric sequences
