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    Let's get right to it.

    "A geometric series has first term 12 and the sum to infinity 18. Find the sum of the first five terms."
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    (Original post by SuperSNESLiam)
    Let's get right to it.

    "A geometric series has first term 12 and the sum to infinity 18. Find the sum of the first five terms."
    What have you tried? Please post your working.
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    Do you know an equation for the sum to infinity of a geometric series? You should be able to use this to find the common ratio, r, and then you'll be able to answer the actual question.


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    (Original post by SuperSNESLiam)
    Let's get right to it.

    "A geometric series has first term 12 and the sum to infinity 18. Find the sum of the first five terms."
    You should be able to write down a and deduce r. You then just need to write down the first 5 terms and add them up (or use the formula).
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    (Original post by SuperSNESLiam)
    Let's get right to it.

    "A geometric series has first term 12 and the sum to infinity 18. Find the sum of the first five terms."
    Take (or derive - it's quick) the expression for the sum of an infinite geometric series (was called a geometric progression in my day), and go from there. The answer will fall-out pretty quickly.
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    (Original post by Pangol)
    Do you know an equation for the sum to infinity of a geometric series? You should be able to use this to find the common ratio, r, and then you'll be able to answer the actual question.


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    ar^0=12

    a/1-r=18

    a=12

    12/1-r=18
    12=18(1-r)
    12=18-r
    -6=-r
    6=r


    Put 12x6^n-1 into my calculator for the first five terms and added them up:

    12+72+432+2592+15,552=18,660.
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    (Original post by SuperSNESLiam)
    Let's get right to it.

    "A geometric series has first term 12 and the sum to infinity 18. Find the sum of the first five terms."
    Extract what you've been given.

     a = 12

     \displaystyle\sum_{n=1}^{\infty} 12r^{n-1} = 18

    Work out what you need to find.

     S_5 = a + ar + ar^2 + ar^3 + ar^4

    Hint: Can you figure out/remember a formula that will give you the value for which this series converges to? Note that  |r|<1 for this series to converge
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    (Original post by SuperSNESLiam)
    ar^0=12
    12=18(1-r)
    12=18-r
    Right idea, but there is a slip between these lines.
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    (Original post by Pangol)
    Right idea, but there is a slip between these lines.
    Ah, I see it.

    12= 18(1-r)
    12=18-18r
    -6=-18r
    1/3=r

    I did 12x1/3^n-1 for the 5 terms and added them to get 484/27.
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    That looks like the right r. Tip for the future - if you've got a geometric series that gives a sum to infinity, then (the size of) r must be less than 1, because otherwise you will be adding terms that are getting bigger every time (or at least not getting smaller). So, you can now see that your first value for r must have been wrong. If this happens in an exam, you can spot this and go back and have another go.
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    (Original post by SuperSNESLiam)
    Ah, I see it.

    12= 18(1-r)
    12=18-18r
    -6=-18r
    1/3=r

    I did 12x1/3^n-1 for the 5 terms and added them to get 484/27.
    Correct. Note that in order for a geometric series to converge to a finite number  L, in other words not  = \infty or = -\infty . -1<r<1 or more simply  |r|<1 . So, if you're given in a question that  S_{\infty} = L where  L is just some finite number then you automatically know that  |r|<1
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    (Original post by Pangol)
    That looks like the right r. Tip for the future - if you've got a geometric series that gives a sum to infinity, then (the size of) r must be less than 1, because otherwise you will be adding terms that are getting bigger every time (or at least not getting smaller). So, you can now see that your first value for r must have been wrong. If this happens in an exam, you can spot this and go back and have another go.
    Thanks. I remember this from the lesson now.

    Thanks guys!
 
 
 
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