$a_n < 0.001$

Algebra Level 3

Let $\{a_n\}$ be a geometric progression such that $a_5=\frac{11}{16} \mbox{ and } a_8=\frac{11}{128}.$ What is the smallest integer $n$ for which $a_n < 0.001?$

13

15

16

14

1 solution

Nick Okita
Mar 31, 2014

The problem gives you a geometric progression of 1/2 ratio, so the first term is 11, and then 11/2, 11/4...

The number you want is smaller than 0.001, therefore it's smaller than 11/(11000). The first 2 power that is bigger than 11000 is approximately 16000, which is 2^14, as the first term is considering 2^0, you have to sum 1 to the power. So it will be n=14+1=15.

how could it be a level three

ashutosh mahapatra - 7 years, 1 month ago

too damn overrated

math man - 6 years, 8 months ago

a n < 0.001 a_n < 0.001 a n ​ < 0 . 0 0 1

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$a_n < 0.001$