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28 December, 22:40

Compute the sums below. (Assume that the terms in the first sum are consecutive terms of an arithmetic sequence.)

9 + 12 + 15 + ... + 336 =

+3
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  1. 29 December, 01:06
    0
    =18975

    Step-by-step explanation:

    This is an arithmetic sequence

    an = a1 + d (n-1)

    a1 = 9

    d = 12-9 = 3

    an = 9 + 3 (n-1)

    The sum of an arithmetic sequence is given by

    Sn = n/2 (a1+an)

    So we need to determine what number in the sequence 336 is

    an = 9 + 3 (n-1)

    336 = 9 + 3 (n-1)

    Subtract 9 from each side

    336-9 = 9+-9 + 3 (n-1)

    327 = 3 (n-1)

    Divide each side by 3

    327/3 = 3/3 (n-1)

    109 = n-1

    Add 1 to each side

    109+1 = n-1+1

    110 = n

    336 is the 110 term

    The sum from 9 to 336

    Sn = n/2 (a1+an)

    S 110 = 110/2 (9+336)

    = 55 (345)

    =18975
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