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Math Help - The sum of certain consecutive odd numbers is 57^2 13^2. Find the numbers.

  1. #1
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    The sum of certain consecutive odd numbers is 57^2 13^2. Find the numbers.

    Problem.
    The sum of certain consecutive odd numbers is 57^2 13^2. Find the numbers.

    solution
    \sum 1+3+5+....+(2n-1) = n^2 [1]

    57^2=n^2

    n=57

    also from equ 1

    n=\frac{l+1}{2}

    therefore
    \frac{l_1+1}{2} = 57

    l_1=113

    similarly
    13^2=n^2

    l_2=25


    57^2 - 13 ^2  = 27^2+29^2+....113^2

    a = 27 , d = 2
    l= 113
    a+nd = 113
    27 + 2n = 113
    n = 43

    this is where I am facing the problem. To satisfy the condition I should have n = 44 and not 43.
    because 57 - 13 = 44. other way to find the n.

    Please correct my understanding...
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  2. #2
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    Quote Originally Posted by chakravarthiponmudi View Post
    a+nd = 113
    The formula for the n'th term of an arithmetic progression is a + (n1)d, not a + nd.
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  3. #3
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    Hello, chakravarthiponmudi!

    You already have all the information you need.


    The sum of certain consecutive odd numbers is 57^2 - 13^2. . Find the numbers.

    Solution

    \sum^n_{k=1}(2k-1) \:=\:1+3+5+ \hdots+(2n-1) \:=\: n^2
    That formula says:
    . . the sum of the first n odd numbers is n^2.

    So 57^2 is the sum of the first 57 odd numbers.
    . . and 13^2 is the sum of the first 13 odd numbers.

    Hence: . 57^2 - 13^2 \:=\:\text{(first 57 odd numbers)} - \text{(first 13 odd numbers})

    . . . . . . . . . . . . = \;(1 + 3 + 5 + \hdots + 113) - (1 + 3 + 5 + \hdots + 25)

    . . . . . . . . . . . . = \;27 + 29 + 31 + \hdots + 113

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