Geometric and arithmetic series?
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Your question (c) should be 3rd term 12 and 6th term 96\n(NOT 16th TERM)\n(6th term)/(3rd term) = 96/12 = 8\n(ar^5)/(ar^2) = 8\nr^3 = 8\nr = 2\n[a = first term, r = common ratio]\n\nYour Question (a) should be 5th term of (2m-k)^9\n[POWER SHOULD BE 9 NOT 4]\nUse binomial formula\n[9!/(5!4!)](2m)^5(-k)^4\n= 126(2^5)(m^5)(k^4)\n= 126(32)(m^5)(k^4)\n= 4032(m^5)(k^4)
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Geometric -- multiplying the previous term by 4
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The ratio (r) between two consecutive terms is 1/3.\n\nJust take two consecutive (neighboring) terms and divide the second by the first.\n\nSo:\n1/3 / 1 = 1/3\n1/9 / 1/3 = 1/3\n1/27 / 1/9 = 1/3\n\nr = 1/3\n\nThis is a very basic question about a geometric series.  Rather than just posting your homework questions here, I suggest learning a bit more about this so you can answer it yourself.
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Geometric series: (termFirst - termLastPlusOne)/(1 - ratio):\n(e^x -e^(7x))/(1-e^x)
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download.com\nsearch for mathtype software and download it!\nall the best!
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2.1 = 21/10^1\n0.21 = 21/10^2\n0.021 = 21/10^3\n0.0021 = 21/10^4\n\nSee the pattern?\n\nIt is a geometric sequence with 1/10 as the common ratio, 21/10 as the first term and it is an infinite series.\n\nUse this formula\n\ns = a1/(1 - r)\ns = (21/10)/(1 - 1/10)\ns = (21/10)/(9/10)\ns = 21/9\ns = 7/3 or 2.333333333333333333333333333333333333333333\n\nHope this helps!!^_^
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hope this helps\n\nhttp://www.thea.nesinc.com/ptmath.htm
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http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP?Res=150\n\nhttp://www.math.sfu.ca/~cbm/aands/subj.htm\n\nhttp://clusterfie.epn.edu.ec/ibernal/html/CURSOS/AbrilAgosto06/Metodos/Libros/Fundamentals%20Of%20Numerical%20Computing%20-%201997%20-Wiley%20%20(By%20Laxxuss).pdf
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Arithmetic means you add a certain number to each term to obtain the next term.\n\nGeometric means you multiply each term by a certain number to obtain the next term.\n\nLook at (A).  You get each term by multiplying the previous term by 1.5, so it's geometric.\n\n(B) is neither.  The additive number changes from term to term, so it's not arithmetic.  No multiplication is involved, so it's not geometric.
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linear...\n\nbecause for a certain values of x, there corresponds only one value of y\n\njust like in arithmetic series
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hav u ever heard of the Principle of Mthematical Induction ?\nthat could solve ur problem in a jiffy!!\n\nHowever if u havn't, dont worry!!\n\nNote that, the sum in ur series(i.e. 1+1/2+1/3+...+1/n),where n>=2, is always less than 2, but greater than 1. so it can't b an integer.\nwanna know why?\ntry cramming up some theorems about Sequence & series of real numbers!! its a regulation part of any under-graduate level Real analysis course.
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umm none that im aware of, what kind of mathamatics leval are you looking at?
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Pick a few easy points like pi/2, pi, 3pi/2, 2pi.  You know the period.  Then just graph it.
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Dear, Your answers with complete explanation are below:\n\n1. Take any 2 terms, say 1/3 and 1/9.\nthe ratio is  = 1/3 / 1/9 = 1/3.\n\nSum of a geometric series to 'n' terms is given by:\nS(n) = a(r^n - 1)/(r - 1),\nwhere, \nS(n) = sum upto n terms\na= first term\nr=ration between 2 consecutive terms\nn= no. of terms\n\n2. Here,\n\nr= 1/3\nn= 10\na = 1.\n\nSo, S(10) = (3^10 - 1)x3 / 3^10 x 2.\nS(10) = 1.49997459 ~ 1.4999746\n\n3. Similarly, S(12) = 1.499997177\n\n4. If there are infinite terms in a geometric series, then its sum is = a/(1-r)\n\nHere this is = 1/(1-1/3) = 3/2 = 1.5\nSo, this series is always less than 1.5.
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http://mathworld.wolfram.com/PoincareConjecture.html
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Check out http://math.mit.edu/research/index.html
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http://mathworld.wolfram.com/SieveofEratosthenes.html