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AP calculus

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Given an arithmetic progression -7,-3,1,..., state three consecutive terms in this progression which sum up to 75.

  • AP calculus - ,

    + 4 = change from term to term

    x-4 + x + x+4 = 75

    3 x = 75

    x = 25
    so
    21, 25, 29

  • AP calculus - ,

    Think about how you can locate these three consecutive terms.

    What is the first term, and the common difference?

    The sum of an AP for the first n terms is given by

    S(n) = (n/2)(2a+(n-1)d)

    where a = -7, d = 4

    So lets say the 3 consecutive terms are the 14,15 and 16th term.

    to get the sum of 75 of these 2 terms, i take

    S(16) - S(13)

    hence,

    S(n) - (n-3) = 75

    substituting the formula,

    (n/2)(2a+(n-1)d) - [(n-3)/2][2a+(n-3-1)d] = 75

    solve and find n.

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