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March 25, 2017

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The 10th term of an arithmetic series is 34, and the sum of the first 20 terms is 710. Determine the 25th term.

  • Math - ,

    Use your formulas
    a+9d = 34 (#1)

    (20/2)[2a + 19d] = 710
    2a + 19d = 71 (#2)

    solve the two equations.
    I would use #1 as a = 34-9d and sub into #2
    Let me know what you got.

  • Math - ,

    The sum of members of a arithmetic progression is:

    Sn=(n/2)*[2a1+(n-1)*d]

    a1-first number in arithmetic progression

    d-common difference of successive members

    n-numbers of members

    In this case:
    n=20 , (n/2)=10 , n-1=19

    a1=a10-9*d
    a1=34-9*d

    Sn=(n/2)*[2a1+(n-1)*d]
    =10*[2*(34-9*d)+19*d]
    =10*(68-18d+19d)=10*(68+d)=680+10d
    Sn=S20
    S20=710
    710=680+10d
    10d=710-680=30
    10d=30 Divided with 10
    d=30/10

    d=3

    a1=a10-9d
    a1=34-9*3
    a1=34-27

    a1=7

    a25=a1+(25-1)*d
    a25=7+24*3
    a25=7+72

    a25=79

    So members of that Arithmetic progression is:

    7,10,13,16,19,22,25,28,31,34,37,40,43,46,49,52,55,58,61,64,67,70,73,76,79

    For more information about Arithmetic progression go to wikipedia and type"Arithmetic progression"

  • Math - ,

    a1=34-9*d
    Becouse nth member in arithmetic progression is:
    an=a1+(n-1)*d

    a10=a1+(10-1)*d
    a10=a1+9*d
    a10-9*d=a1
    a
    a1=a10-9*d
    a1=34-9*d

    an=a1+(n-1)*d
    a25=a1+(25-1)*d
    a25=a1+24*d
    a25=7+24*3
    a25=7+72=79

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