Geometric Series and SequencesDate: 8/9/96 at 20:45:44 From: Anonymous Subject: Geometric Series and Sequences 1. What term of the geometric sequence 3,6,12... is equal to 768? Result so far: Tn = ar^n-1 768 = 3 x 2^n-1 2^n-1 = 256 I'm stuck from here (not even sure if I'm on the right track!). 2. For what range of values for y will the series y+y^2+y^3+....y^n have a limiting sum? Find the limiting sum if y=3/4. 3. If the limiting sum of the geometric series 1+x+x^2+x^3+... is 20, find x. 4. Given the geometric series 2,-6,18,... find the smallest value of n such that |Tn|>3000. 5. Starting with 5, how many consecutive multiples of 5 must be added to make their sum exceed 1000? 6. Express 0.27(repeated) in rational form. (I'm not sure where to start on this one). Sorry to have so many questions, but I'm having real trouble with this. I study by correspondence here in Australia and my tutor is on leave this weekend so I'm really stumped. I'd be very grateful for any assistance - even a starting point. Melanie Date: 8/11/96 at 9:11:34 From: Doctor Anthony Subject: Re: Geometric Series and Sequences Question 1: You are on the right track. Since 256 = 2^8 we have 2^(n-1) = 2^8 This means that n-1 = 8 or n = 9. If you could not see this by inspection, you could use logs to find the value of (n-1). 2^(n-1) = 256 Take ln of both sides: We have (n-1)ln(2) = ln(256) n-1 = ln(256)/ln(2) = 5.545/0.69315 = 8 and so n = 9 as before. Question 2: The geometric series converges if -1 <or= y < 1 The sum to infinity is given by a/(1-r) = (3/4)/(1 - 3/4) = (3/4)/(1/4) = 3 Question 3: S(inf.) = 1/(1-x) = 20 1 = 20 - 20x 20x = 19 and so x = 19/20 Question 4: Here the common ratio is -3 and T(n) = ar^(n-1) > 3000 2*(-3)^(n-1) > 3000 In fact we require 3^(n-1) > 1500 (n-1)ln(3) > ln(1500) n-1 > ln(1500)/ln(3) n-1 > 6.6567 n > 7.6567 n = 8 Question 5: The sequence would be 5 + 5^2 + 5^3 + etc, so this is a GP with common ratio 5 and first term 5. S(n) = a(r^n -1)/(r-1) = 5(5^n - 1)/4 > 1000 5^n -1 > 4000/5 5^n > 801 n*ln(5) > ln(801) n > ln(801)/ln(5) n > 4.154 So n = 5 Question 6: The number is 0.272727.... etc. You can do this in two ways. Probably the easier is to let x = 0.272727.... 100x = 27.272727.... Subtracting 99x = 27 and so x = 27/99 = 3/11 Alternatively, we can consider this as a GP with terms: 27/100 + 27/10000 + etc to infinity with a = 27/100 and r = 1/100 Sum to infinity = a/(1-r) = (27/100)/(1-1/100) = (27/100)/(99/100) = 27/99 = 3/11 as before. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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