lucettafetta Posted November 22, 2014 Report Share Posted November 22, 2014 How do I do this question? u1 = 1/3u2 = 1/9 un = 1/3^n Which is the first term of the sequence that is less than 10^-6 ? (For those who use different notations, u1 = a, and u = t.) Reply Link to post Share on other sites More sharing options...
Thrashmaster Posted November 23, 2014 Report Share Posted November 23, 2014 Well, we clearly see that r=1/3. un is true, so (1/3)^1 is 1/3, (1/3)^2 is 1/9, etc. 10^-6 is 0.000001. Is this a calculator question? If so, it is fairly simple. Just keep typing in (1/3)^n until you get an answer small enough, with n getting bigger each time. My answer is (1/3)^13, so the 13th term. I honestly can't think of a better way to do it, and certainly not a faster one. It'd be ridiculous as non-calc, since it is asking for such a small decimal number that they couldn't possibly expect long division. Is this an IB problem? Anyway, I hope this helped. It might not have... I don't know what you were having trouble with. 1 Reply Link to post Share on other sites More sharing options...
lucettafetta Posted November 23, 2014 Author Report Share Posted November 23, 2014 thanks! That's perfect. Yeah it's a calculator question. Reply Link to post Share on other sites More sharing options...
Vioh Posted November 23, 2014 Report Share Posted November 23, 2014 How do I do this question? u1 = 1/3 u2 = 1/9 un = 1/3^n Which is the first term of the sequence that is less than 10^-6 ? Thrashmaster has answered quite correctly, but it's still trial and error. If the question asks for a much smaller number, then the process of calculating can take a very long time. There's a better trick to do this: Using the information from the question, we know that we have to find a smallest integer 'n' such that it satisfies: 1/3^n < 10^-6 Multiply both sides by 3^n, and then divide both sides by 10^-6, we have: 10^6 < 3^n Take the natural logarithm (or any other log will suffice) of both sides, we have: ln(10^6) < ln(3^n) Using one of the logarithmic rule to take down the power, we have: 6ln(10) < n ln(3) Using basic algebra, we have: 6ln(10) / ln(3) < n Using calculator, we have: 12.6 < n Since 'n' must be an integer that is as smallest as possible, then n = 13 With this method, it'll be really easy to do even when the question asks for a very very small number, for example 10^-100, etc.... Feel free to ask if there's anything unclear. 1 Reply Link to post Share on other sites More sharing options...
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