precalculus

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Find the first six partial sums S1, S2, S3, S4, S5, S6 of the sequence.
2, 5, 8, 11, . . .

  • precalculus -

    The sum of the members of a arithmetic progression :


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


    d = common difference of successive members of arithmetic progression


    In this case :

    S1 = ( 1 / 2 ) * [ 2 * 2 + ( 1 - 1 ) * 3 ]

    S1 = ( 1 / 2 ) * (4 + 0 * 3 )

    S1 = ( 1 / 2 ) * 4 = 2


    S2 = ( 2 / 2 ) * [ 2 * 2 + ( 2 - 1 ) * 3 ]

    S2 = 1 * ( 4 + 1 * 3 )

    S2 = 4 + 3 = 7


    S3 = ( 3 / 2 ) * [ 2 * 2 + ( 3 - 1 ) * 3 ]

    S3 = ( 3 / 2 ) * ( 4 + 2 * 3 )

    S3 = ( 3 / 2 ) * ( 4 + 6 )

    S3 = ( 3 / 2 ) * 10

    S3 = 30 / 2 = 15


    S4 = ( 4 / 2 ) * [ 2 * 2 + ( 4 - 1 ) * 3 ]

    S4 = 2 * ( 4 + 3 * 3 )

    S4 = 2 * ( 4 + 9 )

    S4 = 2 * 13

    S4 = 26


    S5 = ( 5 / 2 ) * [ 2 * 2 + ( 5 - 1 ) * 3 ]

    S5 = ( 5 / 2 ) * ( 4 + 4 * 3 )

    S5 = ( 5 / 2 ) * ( 4 + 12 )

    S5 = ( 5 / 2 ) * 16

    S5 = 80 / 2 = 40


    S6 = ( 6 / 2 ) * [ 2 * 2 + ( 6 - 1 ) * 3 ]

    S6 = 3 * ( 4 + 5 * 3 )

    S6 = 3 * ( 4 + 15 )

    S6 = 3 * 19

    S6 = 57

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