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Math Help - geometric series help asap!!! would be so appppreciated

  1. #1
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    geometric series help asap!!! would be so appppreciated

    please help, i do not understand this question at all:

    a geometric series has first term a and common ratio r where r>1

    the sum of the first n terms of the series is denoted by Sn.

    given that S4= 10 x S2,

    a) find the value of r

    given that S3= 26,

    b)find the value of a,

    c) show that s6 = 728
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  2. #2
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    The formula for a geometric series is Sn=a[(1-r^n)/(1-r)]
    Given that:
    S4=10S2
    S3=26
    r>1 (This is only important to avo1d division by zero)

    Solution:
    Since, S4=10S2 we have:
    a[(1-r^4)/(1-r)]=10[(1-r^2)/(1-r)] multiply by (1-r):
    a[(1-r^4)]=10[(1-r^2)] Factor (1-r^4):
    a[(1-r^2)(1+r^2)]=10[(1-r^2)] divide by (1-r^2):
    a(1+r^2)=10 THIS IS YOUR FIRST EQUATION.
    Next:
    Since S3=26 we have:
    a[(1-r^3)/(1-r)]=26 Long divide 1-r^3 by 1-r to get:
    a(1+r+r^2)=26 THIS IS YOUR SECOND EQUATION.
    Next:
    Divide the first equation by the second:
    (1+r^2)/(1+r+r^2)=10/26 or 5/13. Cross Multiply:
    13+13r^2=5+5r+5r^2 Collect Terms:
    8r^2-5r+8=0
    Which has no real solution thus your problem is impossible
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  3. #3
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    Quote Originally Posted by ThePerfectHacker


    Since, S4=10S2 we have:
    a[(1-r^4)/(1-r)]=10[(1-r^2)/(1-r)] multiply by (1-r):

    S(4)\ =\ 10.S(2) implies:

    a.\frac{1-r^4}{1-r}\ =\ 10.a.\frac{1-r^2}{1-r}

    or, cancelling a and (1-r):

    (1-r^4)\ =\ 10.(1-r^2)

    or:

    \frac{1-r^4}{1-r^2}\ =\ 10.

    Which seems to have solution r\ =\ 3.

    RonL
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  4. #4
    Grand Panjandrum
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    Next:
    Since S(3)\ =\ 26 we have:

    a \frac{1-r^3}{1-r}=26,
    but from the first part we know that r\ =\ 3, so:

    13.a\ =\ 26.
    so a\ =\ 2.

    Finally:

    S(6)\ =\ 2.\frac{1-3^6}{1-3}\ =\ 728.

    RonL
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  5. #5
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    I new I did something wrong I forget the "a" term!
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