SOLUTION: In the geometric sequence 256,128,64,32,…. which is the first term that is less than 0.001?

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Question 1186878: In the geometric sequence 256,128,64,32,…. which is the first term that is less than 0.001?
Answer by ikleyn(52810) About Me  (Show Source):
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In the geometric sequence 256,128,64,32,…. which is the first term that is less than 0.001?
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The progression has 1st term 256 = 2%5E8  and the common ratio r = 1%2F2.


Hence, the n-th term is  a%5Bn%5D = 2%5E8%2A%281%2F2%29%5E%28n-1%29 = 2%5E%288-n%2B1%29 = 2%5E%289-n%29.


We look for the term  a%5Bn%5D < 0.001,  or  2%5E%289-n%29 < 0.001.


Take logarithm base 2 of both sides


    9-n < log%282%2C%280.001%29%29

    9 - n < -9.96...


Since we work with integer numbers, we need to solve

    9 - n <= -10


It is equivalent

    n >= 9 + 10

    n >= 19.


ANSWER.  First term of the GP, which is less than 0.001, is the 19-th term  a%5B19%5D = 256%2A%281%2F2%29%5E18 = 0.000977...   ANSWER

Solved.