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1 May, 07:48

A linear sequence has a common difference of 8. Three consecutive terms in the sequence are added together to give a total of 126 find the first three terms

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Answers (2)
  1. 1 May, 11:23
    0
    Answer: 34, 42, 50

    Step-by-step explanation:

    Let the first term be a

    Second term will be a+d

    Third term will be a+2d

    a+a+d+a+2d = 126

    3a+3d = 126

    Recall that the common difference (d) is 8.

    3a+3d = 126

    3a + (3*8) = 126

    3a+24 = 126

    3a = 126 - 24

    3a = 102

    a = 102/3

    a = 34

    First term is 34

    Second term will be a+d

    = 34+8 = 42

    Third term will be a+2d

    = 34 + (2*8)

    = 34+16

    = 50
  2. 1 May, 11:39
    0
    1st term = a = 34

    2nd term = a + d = 34+8 = 42

    3rd term = a + 2d = 34+2 (8) = 50

    = 34,42,50

    Step-by-step explanation:

    Given;

    Common difference d = 8

    Sum of three consecutive terms = 126

    Using the arithmetic progression AP nth term formula;

    nth term = a + (n-1) d

    Where;

    a = first term

    d = common difference

    So,

    1st term = a

    2nd term = a + d

    3rd term = a + 2d

    Sum of the three consecutive terms;

    a + a+d + a+2d = 3a + 3d

    Equating to the given value;

    3a + 3d = 126

    3a + 3 (8) = 126 (d=8 given)

    3a = 126 - 24

    3a = 102

    a = 102/3 = 34

    Therefore,

    1st term = a = 34

    2nd term = a + d = 34+8 = 42

    3rd term = a + 2d = 34+2 (8) = 50
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