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March 26, 2017

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1.1+2.2+3.3+4.4+5.5.............+100.100=?

  • arithmetic - ,

    Before 100.100 we have 99.99?

  • arithmetic - ,

    this is an arithmetic series, where
    a=1.2
    r=0.1
    n = ? but t(n) = 100.100

    t(n) = a + (n-1)d
    100.100 = 1.1 + .1(n-1)
    99=.1n-.1
    .1n=99.1
    n=991

    so there are 991 terms
    sum(991) = (991/2)(first + last)
    = (991/2)(1.1 + 100.100) = 50144.6

  • arithmetic - ,

    This is not arithmetic series if we have
    ...9.9+10.10+...+99.99+100.100

  • CHECK - arithmetic - ,

    correct, I made a silly error,

    the d = 1.1 instead of 0.1 the way I had it.
    It is an arithmetic series

    new solution:
    t(n) = 1.1 + (n-1)(1.1)
    100.100 = 1.1 + 1.1n - 1.1
    n = 91

    so there are 91 terms
    sum(91) = (91/2)(1.1+100.100) = 4604.6

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