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C2 Geometric

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    In a GP the 2nd term is -12 and the 5th term is 768. Find the common ratio and the first term. how would you do this one?
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    By dividing the terms
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    (Original post by Secreay)
    In a GP the 2nd term is -12 and the 5th term is 768. Find the common ratio and the first term. how would you do this one?
    We know the formula is  \text{nth term} = ar^{n-1}

    Sub in -12 and n=2 in the formula, then sub in 768 and n=5, you will get two equations, solve them simultaneously.
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    -12/r = 768/r^4 seems wrong :/
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    (Original post by Secreay)
    -12/r = 768/r^4 seems wrong :/
     \displaystyle \frac{-12}r = \frac{768}{r^4} \implies \frac{r^4}r = - \frac{768}{12}

    You are going right.
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    but i cant cube root i neg number
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    (Original post by Secreay)
    but i cant cube root i neg number
    why not?
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    (Original post by Secreay)
    but i cant cube root i neg number
    Negative numbers can be cube rooted but you can't take the square root of them. Input it in your calc, and it will give the answer.
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    HAHA im a noob so r=-4?
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    (Original post by Secreay)
    HAHA im a noob so r=-4?
    :yep:
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    (Original post by raheem94)
     \displaystyle \frac{-12}r = \frac{768}{r^4} \implies \frac{r^4}r = - \frac{768}{12}

    You are going right.
    This is how I went around the equation.

    You might find this way easier ^^

    -12r^-1=768r^-4
    -12=768r-4/r^-1
    -12=(768/r^4)/(1/r)
    -12=(768/r^4)*(r/1) [doing the dividing fractions so flip the right]
    -12=768r/r^4 [now apply indices rules here]
    -12=768r^-3
    -1/64=r^-3
    -1/64=1/r^3
    -64/1=r^3/1 [flipped top and bottom]
    -4=r
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    (Original post by Stickyelmo)
    This is how I went around the equation.

    You might find this way easier ^^

    -12r^-1=768r^-4
    -12=768r-4/r^-1
    -12=(768/r^4)/(1/r)
    -12=(768/r^4)*(r/1) [doing the dividing fractions so flip the right]
    -12=768r/r^4 [now apply indices rules here]
    -12=768r^-3
    -1/64=r^-3
    -1/64=1/r^3
    -64/1=r^3/1 [flipped top and bottom]
    -4=r
    I really don't understand, why you quoted me?

    Its not a problem for me to deal with such questions and the way i was doing was easier than your's.

     \displaystyle \frac{r^4}r = - \frac{768}{12}  \implies r^3 = -64 \implies r = \sqrt[3]{-64} = -4

    Who's method is better?
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    (Original post by raheem94)
    I really don't understand, why you quoted me?

    Its not a problem for me to deal with such questions and the way i was doing was easier than your's.

     \displaystyle \frac{r^4}r = - \frac{768}{12}  \implies r^3 = -64 \implies r = \sqrt[3]{-64} = -4

    Who's method is better?
    his method is better....less booky
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    Both methods are exactly the same.. one looks shorter as a lot of the steps can be done mentally and not written down.
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    (Original post by David_Skiller)
    his method is better....less booky
    (Original post by F1Addict)
    Both methods are exactly the same.. one looks shorter as a lot of the steps can be done mentally and not written down.
    Correct, both are the same, but i was surprised to be quoted, may be if the post would have been directed to the OP then it would have made a bit of sense.
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    (Original post by raheem94)
    Correct, both are the same, but i was surprised to be quoted, may be if the post would have been directed to the OP then it would have made a bit of sense.

    "Who's method is better?"

    you were asking the question faget

    go back to doing questions in the book
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    ah woops I quoted the wrong text haha X_X

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