use the geometic sequence of numbers 1,1/2,1/4,1/8... what is the ratio between 2 consecutive terms?
471439
The common ratio is (1/2)/1 = 1/2\nFormula for sum of terms is a(1-r^n)/(1-r) , where a is the first term, r- common ratio and n is the number of terms.\n\nsum of first ten terms is 1(1-(1/2)^10)/(1-1/2) = 1.9980\n\nsum of first twelve terms is (1- (1/2)^12)/(1-1/2) = 1.9995\n\nif n tends to infinity the sum becomes a/1-r  , in this case 1/(1/2) = 2
179181
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!!^_^
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
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.
517585
Starting 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.
916838
(1/2)(1/2)=1/4\n\nB. 1/4
695512
Answers choices are:\n     1, 4, 3, 2, 2,\n      3, 1, 4, 2, 1,\n      1, 1, 4, 2, 1\n      3, 2, 3, 4, 1\n      3, 4, 1, 2, 4
476243
1 1/2\n\nboth a whole number and a fraction
44369
There is an infinite quantity of numbers between 5/8 and 1/4
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It helps to put them in order: 1, 1, 1, 2, 5, 5, 7, 8\nThen count in from the outsides to the middle, and it's halfway between 2 and 5, which is 3.5
575173
thats a geometric sequence with a common ratio of -1... you can get the common ratio by getting the first set of differences... thats 10-12... 12-11.. 11-15.. 15-12... then have the first set of difference: -2 1 -4 3.. then get the second set of difference... -2-1.. 1--4... -4-3... and u get -1 -1.. thats how u get to chek if its a geometric sequence by checking if they have the same second diff... chek the net and search for the formula for solving geometric sequences cuz i forgot bout those already.. goodluck!
804992
Let us call the first 2 "2 sub 1", the second "2 sub 2" ... same thing for the 6's.\n\nLet us say that "2 sub 1" is distinct from "2 sub 2", and so on.\n\nWe have 8 choices for the first number, seven choices for the second number, etc. This means that there are 8*7*6*5*4*3*2*1 choices, or 8!. \n\nNow if you look at any number, take all of the 2's out. You see that you can rearrange the 2's in 3! different ways (including the original arrangement) using the procedure in the paragraph above. If we take out the 2's rearrange them, and put them back in we have created the a new arrangement. So for each arrangement of numbers, we have counted it 3! times. So we divide by 3!. The same thing for the sixes, we divide by 2!.\n\nThe result it (8!)/((3!)(2!)) = 3360 choices.
647980
it would be N+1\n\nBecause a line splits the cirlce in two, two lines in three, etc.
842276
1,2,3,4,6,8,12,16,24,48 \n\nlolz hope this helps :)
971669
1. 1s2 2s2 2p6 3s1\n2. 1s2 2s2 2p6 \n3. 1s2 2s2 2p4\n4. 1s2 2s2 2p6\n5. 1s2 2s2 2p6 3s2 3p6 3d7 4s2
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the 2/4\nyou have to reduce the fractions 2/8 = 1/4 and 2/4  = 1/2
27903
According to a number of on-line dictionarys, a sequence of numbers whose reciprocals form an arithmetic progression:\n\nExamples: 1,2,3,4,... is an arithmetic progression.\n1, 1/2, 1/3, 1/4, ... is a harmonic progression.\n\n1, 11, 21, 31, 41, is an arithmetic progression.\n1, 1/11, 1/21, 1/31, 1/41, ... is a harmonic progression.