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Math Help - Sum of the First 10 Terms

  1. #1
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    Sum of the First 10 Terms

    Find the sum of the first 10 terms if s_2 = 356 and a_8 = 83
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    Quote Originally Posted by sharkman View Post
    Find the sum of the first 10 terms if s_2 = 356 and a_8 = 83
    it would help if we knew what kind of series this is ...
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    Quote Originally Posted by sharkman View Post
    Find the sum of the first 10 terms if s_2 = 356 and a_8 = 83
    What kind of sequence is it? (Arithmetic has a common difference between terms and geometric has a common ratio between terms)
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    Quote Originally Posted by skeeter View Post
    it would help if we knew what kind of series this is ...
    Quote Originally Posted by e^(i*pi) View Post
    What kind of sequence is it? (Arithmetic has a common difference between terms and geometric has a common ratio between terms)

    It's an arithmetic sequence.

    Can you help me now?
    Last edited by mr fantastic; November 16th 2009 at 04:14 AM. Reason: Merged posts
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  5. #5
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    Not a whole lot! My question now is why you have s_2 and a_8? You are giving us one term from each of two sequences!

    Or did your finger just slip on the key board and it is s_2 and s_8 or a_2 and a_8?

    Knowing that a_2= 356 and a_8= 83 tells us that a+ 2d= 356 and a+ 8d= 83. Subtracting the first equation from the second, 6d= 273 so d= 45.5. Now, a+2d= a+ 91= 356 so a= 356- 91= 265. You could, now calculate the first ten terms and add.

    But arthmetic sequences have the very nice property that the "average" of an n consecutive terms is the same as the average of the first and last terms in the sequence. Here, the first term is 265 and the tenth term is [tex]a_{10}= 265- 10(45.5)= 265- 455= -190. The average of those two numbers is (265-190)/2= 75/2= 37.5 and that is the average of the first 10 numbers.
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    see below

    Quote Originally Posted by HallsofIvy View Post
    Not a whole lot! My question now is why you have s_2 and a_8? You are giving us one term from each of two sequences!

    Or did your finger just slip on the key board and it is s_2 and s_8 or a_2 and a_8?

    Knowing that a_2= 356 and a_8= 83 tells us that a+ 2d= 356 and a+ 8d= 83. Subtracting the first equation from the second, 6d= 273 so d= 45.5. Now, a+2d= a+ 91= 356 so a= 356- 91= 265. You could, now calculate the first ten terms and add.

    Are you saying that I now must do the math and calculate the first 10 terms and then add? Can you please show me how this is done?

    But arthmetic sequences have the very nice property that the "average" of an n consecutive terms is the same as the average of the first and last terms in the sequence. Here, the first term is 265 and the tenth term is [tex]a_{10}= 265- 10(45.5)= 265- 455= -190. The average of those two numbers is (265-190)/2= 75/2= 37.5 and that is the average of the first 10 numbers.
    What do you mean by the symbol [tex]a{10}?
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  7. #7
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    I meant a_{10}- your a_10.
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  8. #8
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    I would use the fact that

    U_{n}=a+(n-1)d

    and

    S_{n}=\frac{n}{2}\left[2a+(n-1)d\right]

    where a is the first term and d is the common difference.

    Ideally you should know the proofs behind these.
    you can solve the two linear equations simultaneously for a and d

    356=2a+d

    83=a+7d

    and hence solve for S_{10}
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