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Geometric series homework

  1. Sep 12, 2005 #1
    Hi everyone, I'm new to these forums, so I've only just realized how much help they can be... I have some questions so please, don't hesitate to aid me in my time of need.

    These are regarding geometric sequences and series. I'm supposed to be using S=a+ar^n/1-r where s=the sum of the terms
    a=the first term
    r= the common ratio
    n=the # of terms

    1.) How many generations must a person go back to have at least 1000 ancestors? (The sequence is supposed to go: 2+ 4+ 8+ 16+ ...) I know how to do it, I think, but the part that confuses me is where it says AT LEAST 1000. that means i can't use 1000 in the equation, right? because that's not the sum, that is just approx. the sum. ugh.

    2.) The sum of the first two terms of a geometric series is 12 and the sum of the first 3 terms is 62. Find the first 3 terms of the series. I tried solving the equation for a then substituting it in so i only have one variable, but I got something like, r=r...

    3.) The second term of a geometric series is 15 and the sum of the first 3 terms is 93. Find the first 3 terms of the series. I know this question will be using the same ideas as 2 but once again it doesn't seem to work the way i'm doing it.
    Last edited: Sep 12, 2005
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  3. Sep 12, 2005 #2

    Tom Mattson

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  4. Sep 12, 2005 #3
    Sorry, I know. I did try it. I just edited my post to sort of describe my confusion a little more in depth.
  5. Sep 12, 2005 #4

    Tom Mattson

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    You might have to do some trial and error here. You are looking for the number of terms that gives you the highest sum that is less than 1000. Then, you'll add one to that number of terms.

    From both pieces of info you should be able to get the first term. Once you have that, you should be able to find the second term from the first piece of info. Once you have the first and second terms you should be able to find the ratio between successive terms. I'll let you figure out how to get the third term.

    You are right, this is very much like #2.
  6. Sep 13, 2005 #5


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    For #2, you know the sum of the first 2 terms is 12, and the sum of the first 3 terms is 62, so you can find out the third term.
    Let a_1 be the first term, the second term will be a_1r, the third term will be a_1r ^ 2.
    So you have 2 unknowns : a_1 (the first term), and r (the common ratio).
    And you have 2 equations:
    [tex]\left\{ \begin{array}{l}a_1r ^ 2 = 50 \ (1) \\ a_1(1 + r + r ^ 2) = 62 \ (2)\end{array} \right.[/tex]
    You can do something like:
    Since you have a_1 <> 0
    [tex]\frac{a_1 r^2}{a_1(1 + r + r ^ 2)} = \frac{50}{62} \Rightarrow \frac{6}{31} r ^ 2 - \frac{25}{31} r - \frac{25}{31} = 0[/tex]
    After solving the equation for r, you then substitute r into either equation (1) or (2) to find a_1.
    Or you can solve it in another way, you don't need to first solve for the third term, you have 2 unknowns and 2 equations:
    [tex]\left\{ \begin{array}{l}a_1(1 + r) = 12 \\ a_1(1 + r + r ^ 2) = 62 \end{array} \right.[/tex]
    Solve the two equations, you will have a_1, and r.

    # 3 can be done as # 2.
    Viet Dao,
    Last edited: Sep 13, 2005
  7. Sep 13, 2005 #6


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    There is, by the way, an error in the formula you gave in your first post which may be causing you trouble. The formula for the sum of the first n terms of the geometric series a+ ar+ ... is not "S=a+ar^n/1-r" nor even S= (a+ ar^n)/(1-r).
    It is S= (a- ar^n)/(1- r), subtracting in both numerator and denominator.
    That might be easier to calculate as S= a(1- r^n)/(1-r) or
    [tex]S= a\frac{1-r^n}{1-r}[/tex].

    In the first problem for example, 2+ 4+ 8+ ..., a sum of powers of 2, both a= 2 and r= 2 so [tex]S(n)= 2\frac{1-2^n}{1-2}= 2(2^n-1)= 2^{n+1}- 2[/tex]. Since I remember that 210= 1024> 1000, I would try n+1= 10.
    Last edited by a moderator: Sep 13, 2005
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