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Today, 02:30

If the second term of an arithmetic sequence is 5 and the fourth term is 12, what is the 37th term?

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  1. Today, 04:34
    0
    Aritmetic sequence

    an=a1+d (n-1)

    a1=first term

    d=common differnce

    an=nth term

    given

    a2=5

    a4=12

    so

    a2=a1+d (2-1) = 5

    a4=a1+d (4-1) = 12

    a1+d=5

    a1+3d=12

    so 12-5 = (a1+d) - (a1+3d) = 2d=7

    2d=7

    divide by 2

    d=3.5

    a2=a1+3.5 (1)

    5=a1+3.5

    minus 3.5 both sides

    1.5=a1

    an=1.5+3.5 (n-1)

    37th term

    a37=1.5+3.5 (37-1)

    a37=1.5+3.5 (36)

    a37=1.5+126

    a37=127.5

    37th term is 127.5
  2. Today, 06:20
    0
    By using the information in the question, find a equation in the y=mx+b form. Then, plug in 37 as the x value and solve for y.
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