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Math Help - sum of an arithmetic progression

  1. #1
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    sum of an arithmetic progression

    The ratio of the 5th term to the 12th term of a sequence in an arithmetic progression is 6/13. If each of this sequence is positive, and the product of the 1rst term and the 3rd term is 32, find the sum of the first 100 terms of this sequence.

    Hope someone can help, I'm studying for my math final. Which formula should I use?
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    You are told that \dfrac{U_5}{U_{12}} = \dfrac{6}{13} and U_1U_3 = 32.

    By definition a = U_1 and U_n = a+(n-1)d

    Next work out U_5 and U_{12} and U_3 in terms of a and d and the solve the simultaneous equations

    Edit: Sum of an arithmetic sequence is S_n = \dfrac{n}{2}\left(2a+(n-1)d\right)
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    Before finding \displaystyle S_{100} you need to find  \displaystyle a and \displaystyle d such that

    \displaystyle\frac{t_5}{t_{13}}=\frac{6}{13}=\frac  {a+(5-1)d}{a+(13-1)d}

    and

     \displaystyle t_1\times t_3 = t_3 \times a = 32

    Does this help?
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    Hey, thank you so much for helping me but I'm still not getting it, could you please explain further?
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    Quote Originally Posted by stressedIBstudent1 View Post
    Hey, thank you so much for helping me but I'm still not getting it, could you please explain further?
    You should know that U_5 = a+4d and U_{12} = a + 11d

    \dfrac{a+4d}{a+11d} = \dfrac{6}{13} \text{..........  eq(1) }

    U_1U_3 = a(a+2d) = 32 \text{..........  eq(2) }


    Solve simultaenously for a and d.
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    Quote Originally Posted by stressedIBstudent1 View Post
    The ratio of the 5th term to the 12th term of a sequence in an arithmetic progression is 6/13. If each of this sequence is positive, and the product of the 1rst term and the 3rd term is 32, find the sum of the first 100 terms of this sequence.

    Hope someone can help, I'm studying for my math final. Which formula should I use?
    T_1=a,\;\;T_2=a+d,\;\;T_3=a+2d,\;\;T_4=a+3d,\;\;T_  5=a+4d,\;\;.....T_{12}=a+11d

    \displaystyle\frac{T_5}{T_{12}}=\frac{a+4d}{a+11d}  =\frac{6}{13}\Rightarrow\ 13(a+4d)=6(a+11d)

    \Rightarrow\ 13a+52d=6a+66d\Rightarrow\ 13a-6a=66d-52d\Rightarrow\ 7a=14d\Rightarrow\ a=2d

    Substitute this "a" into a(a+2d)=32


    Alternatively work the other way...

    32=1(32)=2(16)=4(8) so you could test a=2,\;\; d=7 and a=4,\;\;d=2 in the ratio equation.
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