the second term of a geometrical sequence is 2, and the fifth is 250. what is the seventh term?
140903
6250\n\nX    2    10    50    250   1250    6250\n\nEach term is 5x the previous.
198100
7 / [8 2/5] = 7 / [42/5] = 7* 5/42= [7*5]/[7*6] = 5/6
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.
755601
26. The pattern has to start 1, 2, 4, 8, 15. I see the differences are 1, 2, 4, 7, and the differences of the differences are 1, 2, 3. The next difference of the difference has to be 4, so, working my way back up, I figure the next difference has to be 11, making the next number 26.\n\nAs for why this makes sense: Consider the question in two dimensions, i.e. into how many parts will many random lines diivide a plane?  That pattern starts 1, 2, 4, 7, 11 - the same numbers on the differences line above.  If you want to get really abstruse, I would submit that the number parts random spaces divide the fourth dimension is 1, 2, 4, 8, 16, 31, 57.  :)\n\n* * * * * *\n\nHey, you really do ask an interesting question, but it appears I'm the only one willing to answer it.  (Yes, I see you asked the same question a few days ago.)  OK, then, let's look at it more systematically:\n\n\nOne-dimensional question: Into how many parts will random points divide a line?\n\nEasy question, obviously:  Zero points leave one part (i.e. the whole line); one point breaks it into two parts, two points into three parts, etc.  The sequence is 1, 2, 3, 4, 5, etc.  Generally, n random points divide a line into n + 1 parts. \n\n\nTwo-dimensional question: Into how many parts will random lines divide a plane?\n\nHarder question, but you can work this one out on paper.  Zero lines leave one part (the whole plane); one line breaks it in two; two lines breaks it in four.  Then it gets tricky - the third line can't go thru all four parts, but it can go through three of them, so you end up with 7 parts.  The fourth line can go through 4 of the 7 parts.  Now, how you'd prove this rigorously I don't know, but the sequence is 1, 2, 4, 7, 11, 16, etc. and is clearly derived from the one-dimensional sequence: start with 1, then add 1 to get 2, then add 2 to get 4, then add 3 to get 7, then add 4 to get 11, etc.  To put this into a formula, n random lines divide a plane into (n^2 + n + 2)/2 parts.        \n\n\nNow for the three-dimensional question: Into how many parts will random planes divide a three-dimensional space?\n\nTo visualize this one, look at a corner of your room.  Obviously, zero planes leave one part (the whole space), one plane  breaks it in two (visualize the floor of your room as the plane, for example), two planes breaks it in four (visualize the floor plus one wall), and three planes breaks it in eight (visualize the floor plus two adjoining walls).  Now, try to imagine the fourth plane - it can't go through all eight parts, but it can go through seven of them, making a total of 15 parts.  The three-dimensional  sequence starts 1, 2, 4, 8, 15, and again we see it is derived from the sequence for the previous dimension: start with 1, then add 1 to get 2, then add 2 to get 4, then add 4 to get 8, then add 7 to get 15, etc.  To get the number for 5 planes, add 11 to get 26.  To put this into a formula, n random planes divide space into (n^3 + 5n + 6)/6 parts.   \n\nBased on what we are seeing, what the heck - why not carry it forward to the fourth dimension.  Using the numbers from the three-dimensional sequence, the sequence becomes 1, 2, 4, 8, 16, 31, 57, etc.  We can conclude, for example, that 6 random three-dimensional spaces divide the fourth dimension in 57 parts.  Now that should really impress your instructor!
594357
You asked for the nth term, so here it is\n\nFigure: n\n\nSquares: n(n+1)/2\n\nDots:  (n^2 + 5n + 2)/2\n\neg. n = 5\n\nsquares = 5 x 6/2 = 15\n\ndots: (25 + 25 + 2)/2 = 52/2 = 26
786690
u can do it by\n(n/2)(a+l)\nwhere \nn is no. of terms\na is first term\nl is last term
368197
6467.\n\nIf u write the numbers as follows:\n64 67\n94 55\n75 68\n31 24\n93 64\n\n6467 is the only number in which the second two digit no. (67) is larger than the first (64).
893652
with 1q, 2 ways (0,1 n)\nwith 3d, 1 ways (0 n)\nwith 2d, 3 ways (0,1,2 n)\nwith 1d, 5 ways (0,1,2,3,4 n)\nwith 0d, 7 ways (0,1,2,3,4,5,6 n)\n\nTotal of 18 ways
1167473
Here is the answer to your question (using the Pythagorean Theorem):  a^2 + b^2 = c^2, a and b are legs of the triangle, c is the hypotenuse.  So c = 26.\n\nso you have:  a^2 + b^2 = c^2\n                       a^2 + b^2 = 26^2\n                       a^2 + b^2 = 676\n                       \nNow, you need two squares that add up to 676.  The only squares that will work are 10 and 24.  So, here is what it looks like:\n\n                     10^2 + 24^2 = 26^2\n                     100 + 576 = 676\n                          676 = 676\nSince you get the same thing on both sides, you have the right answer.  This means that 10 and 24 are the lenglths of the other two sides of the triangle.\n\nHope this answers your question.\n\nChris (Math Teacher)
881801
... 27, 26, 676, 678, 677, 458329, 458331, 458330, ...\n\nThis is sequence A059417 in the Online Encyclopedia of Integer Sequences.\n\nSquare, Add 2, Subtract 1, repeat...
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!
248891
The sum of the first n terms of a geometric series is a1((1-r^n)/(1-r))\n\nwhere a1 = the first term and r = the common ratio.  Here, a1 = 7 and r = 1/3 (because you multiply each term by 1/3 to get the next one).\n\nSo the sum is 7((1-(1/3)^7)/(1-1/3))\n\n= 7(2186/2187)/(2/3)\n\n= 10.4951989 or 10 and 361/729
218316
Thats funny.  Tell me another one.\n\nMaybe its like 5-2=2\n\n\nlol
1006785
the correct sequence is:-\n\nthousand,million,billion,Trillion, Quadrillion, Quintillion, Sextillion, Sept, Oct, non, deci .....\ni dont know more than that.
527221
The answer is 2x+7 with a remainder of 32
627773
1,2,5,10,20,50,100.\n\nSo that's 7.
124205
7, 7, 2\n\n7*7=49\n49*2 = 98