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Math Help - Stuck on progressions question

  1. #1
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    Stuck on progressions question

    How many terms of the series 4,10,16, ....... must be taken such that the sum is equal to 602?

    So far i have:

    Sn=n/2(2a+(n-1)d)
    602x2=n(2x4+(n-1)6)
    1204=n(8+6n-6)
    1204=8n+6n squared-6n
    1204=6n squared+2n
    602=3n squared+n
    n+3n squared -602=0

    I have got my self into a bit of a pickle with this now and am stuck any help would be much appreciated

    cheers
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  2. #2
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    3n^{2}+n-602 = 0

    You can solve the equation by factorization

    Notice -602 X 3 = -1806 = 43X-42

    3n^{2}+43n-42n-602 = 0

    Obtain the roots of this equation thus, you'll get the answer.

    If you're having trouble factorizing, you could use the formula for the roots of a quadratic equation ax^{2}+bx+c = 0

     x=(-b-(b^{2}-4ac)^{1/2})/2a and  x=(-b+(b^{2}-4ac)^{1/2})/2a
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  3. #3
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    Hello, boboulas!

    You're work is correct . . .


    How many terms of the series 4,10,16 . . .must be taken
    . . such that the sum is equal to 602?

    So far i have:

    S_n\:=\:\frac{n}{2}[2a+(n-1)d]

    602\cdot2\:=\:n[2\cdot4+(n-1)6]

    1204\:=\:n(8+6n-6)

    1204\:=\:8n+6n^2-6n

    1204\:=\:6n^2+2n

    602\:=\:3n^2 +n

    n+3n^2-602\:=\:0

    I have got myself into a bit of a pickle . . .

    You have a quadratic equation: . 3n^2 + n - 602 \:=\:0

    Factor (or use the Quadratic Formula): . (n-14)(3n+43) \:=\:0

    . . And we have: . n \:=\:14,\:-\frac{43}{3}


    Since \,n must be a positive integer: . n = 14

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