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    The question is, the Geometric series has common ratio of 5, and the sum of the first three terms is 37.2, Find the first term.

    Im kinda confused I put it in the nth term formula, and at first I thought it would be

    A x 5 to the power of 2 = 37.2 but thats doesnt get me to the right answer.
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    (Original post by SunDun111)
    The question is, the Geometric series has common ratio of 5, and the sum of the first three terms is 37.2, Find the first term.

    Im kinda confused I put it in the nth term formula, and at first I thought it would be

    A x 5 to the power of 2 = 37.2 but thats doesnt get me to the right answer.
    I've emphasised the important bit. The nth term formula is not the sum of the first three terms. The sum of the first three terms is A + 5A + 25A.
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    (Original post by Zacken)
    I've emphasised the important bit. The nth term formula is not the sum of the first three terms. The sum of the first three terms is A + 5A + 25A.
    Yeah I got the answer, another question on it, I think I applied what you told me, This time I had to find the Common Ratio when the Sum of the first three terms is 129 and the frist term is 3,

    So would I write it as 3R to the power 0 + 3R + 3R to the power of 2 = 129?
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    Yeah
    3+3r+3r^2 = 129
    Then form a quadratic equation to get 3r^2+3r-126=0 Simplify to r^2+r-42=0
    (r+7)(r-6)=0
    So r= (-7) or 6
    r=6
    Also Dont forget to check this by putting 3+3(6)+3(36) in your calculator
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    (Original post by SunDun111)
    Yeah I got the answer, another question on it, I think I applied what you told me, This time I had to find the Common Ratio when the Sum of the first three terms is 129 and the frist term is 3,

    So would I write it as 3R to the power 0 + 3R + 3R to the power of 2 = 129?
    Huh? The first term is 3 and yet you've written 0 + ...?

    If the first term is 3, then the second term is 3r and the third term is (3r)r = 3r^2.

    Hence the sum of the three terms is 3 + 3r + 3r^2 = 129. Which bit confuses you here?
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    (Original post by Marshall1999)
    Yeah
    3+3r+3r^2 = 129
    Then form a quadratic equation to get 3r^2+3r-126=0 Simplify to r^2+r-42=0
    (r+7)(r-6)=0
    So r= (-7) or 6
    r=6
    Also Dont forget to check this by putting 3+3(6)+3(36) in your calculator
    I've bolded the incorrect bits in your answer. Also, is there a reason why you're rejecting r=-7? Is there some condition that I'm overlooking?
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    (Original post by Zacken)
    Huh? The first term is 3 and yet you've written 0 + ...?

    If the first term is 3, then the second term is 3r and the third term is (3r)r = 3r^2.

    Hence the sum of the three terms is 3 + 3r + 3r^2 = 129. Which bit confuses you here?
    No I meant to put 3 not sure why I put 0 haha, thanks I got that! I was just checking to see if I was doing it right
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    (Original post by SunDun111)
    No I meant to put 3 not sure why I put 0 haha, thanks I got that! I was just checking to see if I was doing it right
    Ah, okay. Sounds good, then!
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    (Original post by Zacken)
    I've bolded the incorrect bits in your answer. Also, is there a reason why you're rejecting r=-7? Is there some condition that I'm overlooking?
    Thanks. no id just done a question where r had to be positive so i got a little confused
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    (Original post by Marshall1999)
    Thanks. no id just done a question where r had to be positive so i got a little confused
    Fair enough.
 
 
 
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