Geometric Sequences and Exponential Functions

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Key Questions

  • A geometric sequence is given by a starting number, and a common ratio.

    Each number of the sequence is given by multipling the previous one for the common ratio.

    Let's say that your starting point is #2#, and the common ratio is #3#. This means that the first number of the sequence, #a_0#, is 2. The next one, #a_1#, will be #2 \times 3=6#. In general, we have that #a_n=3a_{n-1}#.

    If the starting point is #a#, and the ratio is #r#, we have that the generic element is given by #a_n=ar^n#. This means that we have several cases:

    1. If #r=1#, the sequence is constantly equal to #a#;
    2. If #r=-1#, the sequence is alternatively equal to #a# and #-a#;
    3. If #r>1#, the sequence grows exponentially to infinity;
    4. If #r<-1#, the sequence grows to infinity, assuming alternatively positive and negative values;
    5. If #-1<r<1#, the sequence exponentially decreases to zero;
    6. If #r=0#, the sequence is constantly zero, from the second term on.
  • It means that the next term can always be obtained by multiplying a current term by #r#.

    #a_1=a#

    by multiplying by #r#,

    #a_2=ar#

    by multiplying by #r#,

    #a_3=ar^2#
    .
    .
    .


    I hope that this was helpful.

  • A geometric sequence is always of the form
    #t_n=t_"n-1"*r#

    Every next term is #r# times as large as the one before.

    So starting with #t_0# (the "start term") we get:
    #t_1=r*t_0#
    #t_2=r*t_1=r*r*t_0=r^2*t_0#
    ......
    #t_n=r^n*t_0#

    Answer:
    #t_n=r^n*t_0#
    #t_0# being the start term, #r# being the ratio

    Extra:
    If #r>1# then the sequence is said to be increasing
    if #r=1# then all numbers in the sequence are the same
    If #r<1# then the sequence is said to be decreasing ,
    and a total sum may be calculated for an infinite sequence:
    sum #sum=t_0/(1-r)#

    Example :
    The sequence #1,1/2,1/4,1/8...#
    Here the #t_0=1# and the ratio #r=1/2#
    Total sum of this infinite sequence:
    #sum=t_0/(1-r)=1/(1-1/2)=2#

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