Question 1186878
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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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<pre>
The progression has 1st term 256 = {{{2^8}}}  and the common ratio r = {{{1/2}}}.


Hence, the n-th term is  {{{a[n]}}} = {{{2^8*(1/2)^(n-1)}}} = {{{2^(8-n+1)}}} = {{{2^(9-n)}}}.


We look for the term  {{{a[n]}}} < 0.001,  or  {{{2^(9-n)}}} < 0.001.


Take logarithm base 2 of both sides


    9-n < {{{log(2,(0.001))}}}

    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.


<U>ANSWER</U>.  First term of the GP, which is less than 0.001, is the 19-th term  {{{a[19]}}} = {{{256*(1/2)^18}}} = 0.000977...   <U>ANSWER</U>
</pre>

Solved.