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22 March, 11:46

What is the fifth term of the geometric sequence a1=120, a2=36, a3=10.8, a6=0.2916?

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  1. 22 March, 11:51
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    The formula for geometric sequence is an = a1 r^ (n-1) where r is the geometric factor and n is an interger. In the sequence given, r is equal to 3/10. In this case, an = 120 * (3/10) ^ (n-1). Solving for a5, a5 = 120 * (3/10) ^ (5-1) = 0.972
  2. 22 March, 11:58
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    First we need to find the common ratio by dividing the second term by the first term. 36/120 = 3/10

    an = a1 * r^ (n-1)

    n = term to find = 5

    a1 = first term = 120

    r = common ratio = 3/10

    now we sub and solve

    a5 = 120 * 3/10^ (5 - 1)

    a5 = 120 * 3/10^4

    a5 = 120 *.0081

    a5 = 0.972 < = = = fifth term

    or we could have just done this ... since we multiply by 3/10 to find the next number ...

    a3 = 10.8 ... 10.8 * 3/10 = 3.24 < = = 4th term

    3.24 * 3/10 = 0.972 < = = fifth term
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