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Math Help - sum to infinity - GP

  1. #1
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    Post sum to infinity - GP

    i cant understand this question... because i was away when they learn it so yea...

    Find the 1st 3 terms of a geometric sequence, given that the sum of the 1st three terms is 76/3 and the sum to infinity is 36.
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  2. #2
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    GP

    Hello erisse_satg17
    Quote Originally Posted by erisse_satg17 View Post
    i cant understand this question... because i was away when they learn it so yea...

    Find the 1st 3 terms of a geometric sequence, given that the sum of the 1st three terms is 76/3 and the sum to infinity is 36.
    Welcome to Math Help Forum.

    This is what you need to know. In a GP:

    • the first term is denoted by a
    • the common ratio is denoted by r
    • the first three terms are a, ar, ar^2
    • the sum of the first n terms is \frac{a(1-r^n)}{1-r}
    • the sum to infinity is \frac{a}{1-r}, provided |r| < 1

    So you'll need to plug in the values you've been given:

    \frac{a(1-r^3)}{1-r}=\frac{76}{3}

    and \frac{a}{1-r}=36

    and solve for a and r. (Don't forget to answer the question when you done this.)

    Can you complete it now?

    Grandad
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  3. #3
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    thanks for your help...

    so im gonna use simultaneous to get the answer ryt?
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  4. #4
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    Hello erisse_satg17
    Quote Originally Posted by erisse_satg17 View Post
    thanks for your help...

    so im gonna use simultaneous to get the answer ryt?
    Yes - and it's very easy. You can just get rid of the \frac{a}{1-r} in the first equation, by using the second equation to replace it by 36. Like this:

    \frac{\color{red}a\color{black}(1-r^3)}{\color{red}1-r}=\frac{76}{3}

    \Rightarrow \color{red}36\color{black}(1-r^3)=\frac{76}{3}

    Can you finish it now?

    Grandad
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