find the first term of the sequence 24,8,8/3,8/9.....which i less than 0.001?
108578
the first term is A11=8/3^9\nAn = A1*q^(n-1)\nin our sequence A1=24 and q=3\ntherefore An=24*(1/3)^(n-1)\nlet's say 0.001 = 24*(1/3)^(n-1)\n0.001/24 = (1/3)^(n-1)\n1/24000 = ((1/3)^n)/(1/3)        /:1/3\n1/72000 = (1/3)^n\nn = (log(1/72000))/(log(1/3))\nn =  10.18\nwe want An to be less then 0.001 so we take a greater integer n, so n = 11 and A11 = 24*(1/3)^10 = 8/3^9
243446
It's called the Fibonacci sequence, but there's actually a couple more terms at the front: 0,1,1,2,3,5,8,13...  You take the last 2 terms and add them together to get the next term.  So 8+13 = 21; 21+13 = 34, etc.\n\nHere is a good start if you're interested in learning more about it:\n\nhttp://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/
1040147
I'm guessing you mean 21/24 = x / 8\nthat's 7/8 in lowest terms\n\nx = 7
158511
3/8+2/3-3/4=(9+16-18)/24=7/24
611894
8 to 3  (56 to 21 and divide by 7)
1103012
35\n\nThe sequence is every even-indexed Fibbonacci number plus 1.  So from the original sequence whose value is the sum of the two previous numbers:\n\n0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...\n\nYou take every second number and add 1:\n\n2, 3, 6, 14, 35, 90, 234, ...
413378
24, factors are 1, 2, 3, 4, 6, 8, 12, 24
943548
36\n\nFirst time= x-(x/2+2) = x/2-2\nSecond time=x/2-2-(1/2(x/2-2)+2)=x/4-3\nThird time=x/4-3-(1/2(x/4-3)+2)=x/8-3.5\nx/8-3.5=1
868097
-\naside from 1,\n2,3,4,6,12
887960
You're answer is perfectly correct.  You do want to use root test, and you applied it properly.\n\nYou could also check your work by sorta "thinking" it out.\n\nthe first term (if you start with n=1) is 1, then 1/4 then 1/27 then 1/256 and so on, so as you can see, if you sum the first 3 terms we get a number close to 1.3, if we sum the first 4 numbers its still about 1.3... doesn't matter how many terms it is, i think it ends up around 1.3 or 1.33333 or something, there is a way to figure out what the sum converges to.
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, .....
723065
The sequence is the given series is *2 *3 *4 *5 *6and so on...so the answer is 720
1006179
111221, 312211, 13112221, 1113213211, ...\nOR\n1231, 131221, 132231, 232221, 134211, ...\nOR\n3112, 211213, 312213, 212223, 114213, ...\n\nEach is based on the digit makeup of the previous term, but I need one more term in the sequence to be sure of the rule.
619316
Do you have the same homework as Tilo B?\n\nStarting with 0, the sequence is the square of that number:\n\nf(n) = n^2. \n\nf(0) = 0 (0^2)\nf(1) = 1 (1^2)\nf(2) = 4 (2^2)\nf(3) = 9 (3^2)\nf(4) = 16 (4^2)\nf(5) = 25 (5^2)\netc.\n\nNote, it isn't clear if your indexes are going 0,1,2,3,4,... or are going in descending order 0,-1,-2,-3,-4... the resulting sequence would be the same.\n\nf(0) = 0 (0^2)\nf(-1) = 1 (-1^2)\nf(-2) = 4 (-2^2)\nf(-3) = 9 (-3^2)\nf(-4) = 16 (-4^2)\nf(-5) = 25 (-5^2)\netc.
731226
Turn them into improper fractions:\n47/8+41/12\n141/24+82/24=\n223/24 or 9 7/24
1051900
1 2 3 4 5 6 10 12 15 20 30 60\n-1 -2 -3 -4 -5 -6 -10 -12 -15 -20 -30 -60
1032174
Observe:\nsum= 4+12+24+40+60+.....+t(n-1)+t(n)\nsum=       4+12+24+40+.....+t(n-2)+t(n-1)+t(n)\nSubtracting,\nt(n)=4+8+12+16+20+.........upto n terms\n      = n/2[2*4+(n-1)*4]\n      = n/2[4+4n]=n[2+2n]