Help finding the following arithmetic sequence ...plz =)?
925404
OK, let's see.\n\nIt's an arithmetic sequence, so the difference between consecutive terms is a constant.  The first term is 1, so we can express each term as:\n\nx(i) = 1 + k*(i-1), i = 1, 2, 3, ....\n\nNext the problem says that the 2nd, 10th, and 34th terms are the first three terms in a geometric sequence.  That means that the ratio of the 10th term to the 2nd term must be the same as the ratio of the 34th term to the 10th, because when you divide any two terms of a geometric sequence, you get the same constant.\n\ni(2) = 1 + 1*k\ni(10) = 1 + 9*k\ni(34) = 1 + 33*k\n\nSo we said i(10)/i(2) must equal i(34)/i(10):\n\n1+9k    1+33k\n-------- = --------\n1+1k      1+9k\n\nCross multiplying: (1+9k)^2 = (1+1k)(1+33k).\n\nThen 1 + 18k + 81k^2 = 1 + 34k + 33k^2.\n\nPutting it all on the left side, 48k^2 - 16k = 0, or 16k(3k - 1) = 0.\n\nSo either k = 0 (not interesting) or k = 1/3.\n\nTherefore the arithmetic sequence must be:\n\n1, 1 1/3, 1 2/3, 2, 2 1/3, 2 2/3, 3, 3 1/3, ....\n\nThe second term is 1 1/3.  The 10th term is 4 (1 + 3 * 1/3).  The 34th term is 12 (1 + 33 * 1/3).  So the geometric progression would be:\n\n4/3, 4, 12, 48, 192, .....
394920
1) [1/x] <= 1/x, for all x>0\n2) [1/x] = 1/x, iff 1/x is an integer, so x=1/n for some integer n.\n3) [1/x] < 1/x, for all x != 1/n for any n (x>0).\n4) So, f(x) = [x*[1/x]] = [x*1/x] = [1] = 1 for all x=1/n for some integer n.\n5) Also, x*[1/x] < x*1/x = 1 for all x != 1/n for any n (x>0),\nthen f(x) = [x*[1/x]] = [<1] = 0 for all x != 1/n for any n (x>0).\n\nlim f(x)(x->a) = b, iff for ALL sequences {a_n->a} f(a_n)->b.\n\nLet's take 2 sequences:\n{a_n = 1/n} f(a_n)=1 for all n (see 4), so f(a_n)->1.\n{b_n = 2/(2*n+1)} f(b_n)=0 for all n (see 5), so f(b_n)->0.\nEasy to see that {a_n}->0, {b_n}->0.\n\nThus, there is NO limit of f(x) when x->0.\n\nPS: "iff" means "if and only if"
286646
Try this: \nhttp://www.research.att.com/~njas/sequences/?q=3%2C3%2C4%2C5%2C7%2C8&sort=0&fmt=0&language=english&go=Search\n\nthere are several possible solutions, see what matches what you wanted
503153
(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too
1030368
(a1* ... *aN)^(1/N)\n\nAs soon as you find the geometric sequence\n\n(a1 * a2 * a3 * a4 * a5 * a6)^(1/6)\nwould be an example of geometric meant, assuming you know a1 through a6\n\nLets say you had\n\n1, 2, 4, 8, 16, 32\n\nThe pattern would be 2^(n - 1)\n\nThe geometric mean would be done like this\n\n(1 * 2 * 4 * 8 * 16 * 32)^(1/6) = 32768^(1/6) = 5.6568542...
889891
~..-8,-6,-4,-2,0,2,4,6,8
909352
This question has been asked so many times!?!?!\n\nGood challenge question!\n\nIt's a tough one, too.\n\nI'm taking it to work with me and asking my colleges to help me out on it.\n\nSomeone "did" crack this CHALLENGE Question.\n\nSo here's the answer. . .\n\nhttp://answers.yahoo.com/question/index;_ylt=AvnsMpyNhtwrHV1JVMvXrE4jzKIX?qid=20061130045858AAB57A7\n\nANSWER:  ABCDEFGHJ = 25,693,148\n\nA=2\nB=5 \nC=6\nD=9\nE=7\nF=3\nG=1\nH=4\nJ=8
129538
The next number is 30651.\nIt has a recursion, a_n=a_(n-1)^2+2*a_(n-2).
287856
This seems not a geometric sequence:\n\nU(1) = 6*(1/2)^1-1 = 2.0\nU(2) = 6*(1/2)^2-1 = 0.5\nU(3) = 6*(1/2)^3-1 = -0.25\n\nU(3) / U(2) is not the same as U(2) / U(1)\n\nDo you mean geometric sequence or geometric series?\n\n------- more note ---------\n\nThe answer by another user below indicated that the formula was: U(n) = 6*(1/2)^(n-1).\n\nSomeone needs to learn more about math operators and their precedence :-).
136789
That's an extremely good question. \n\nThe answer is simple. Before that, i will explain what your question means to those who don't get. Well, it means that we must get 24 using 1,3,4 and 6 any number of times however involving operations +,-,* and / only once. \n\n  Then, the answer is very simple :\n\n  {[(6*6)/(3-1)]+[6]}  =  24\n\n---------------------------------------------------------\nWOKRING :\n\n {[(6*6)/(3-1)]+[6]}   =  {[(36)/(2)]+[6]}\n                       =  18 + 6\n                       =  24\n\n----------------------------------------------------------\n\nI hope you post more questions like these. They evoke a lot of thought. I hope the solution serves th purpose.
280927
let the first 3 digits of your number be x = abc and the last four digits be y =defg\n\nwhere a-g are the digits\n\nThe sequence of operations can be rewritten as:\n\n((x*80 + 1) * 250 + 2 * y - 250)/2\n\nIf you distribute the first 250 you get\n\n(20000x + 250 + 2y - 250)/2\n\nCancel the 250's\n\n(20000x+2y)/2\n\nDivide both elements of the numerator by two (distribute the 2)\n\n10000x + y\n\nSince x is a three digit number abc, abc * 10000 = abc0000\n\nSince y is a four digit number defg, 10000x + y = abcdefg.  \nThus we see using algebra that you end up with the same 7 digit number you started out with.
911176
Hi Shwanta  \n\nIf that number we are looking for is " x " \nso we have ;\n1/3 x + 5 \n\nGood Luck dear
821513
c.  It's the 7th term\n\nHere is the entire sequence:\n3/16, 3/8, 3/4, 3/2, 3/1, 3/(1/2), 3/(1/4).  Each time you take half of the denominator.  3/(1/4) = 3*4=12.
810843
I am not too clear on what you are asking for, but I will attempt two options.\n\nA monotonic sequence that ends abruptly could be halving of natural numbers.  For example, starting off with 128, we obtain 64,32,16,8,4,2,1,0.  (The integer division of 1 by 2 is 0).\n\nIf you are talking about a finite sum, then it could also be the sum of numbers that keeps halving, such as\n\n1, 0.5, 0.25, 0.125, 0.0625, 0.03125, 0.015625, 0.0078125....\nThe sum to infinite terms is 2.
432827
(n-1)!/(n+1)! = (n-1)!/((n+1)n(n-1)!)\n= 1/((n+1)n)\n= 1/(n^2+n)
727389
1) 21\n2) 52\n3) 40\n4) 80\n5) 104\nIn each pattern you are adding a number to the one before it.\nEx: 1 + 0 = 1, 1 + 1= 2, 2 + 1 = 3, 3 + 2 =5, 5 + 3 = 8, etc.
889816
1/rt3 *1/rt2-1/rt3\n[1/rt3](1/rt2-1)\n[1/rt3](1-rt2/rt2)\nrationalising\n[(1-rt2)*rt6]/6