I have a math question on geometric sequences.  How do u solve the following?
548552
If you mean "to" 10%, the answer is:\n\n.64^n = .10\n\nn * (log .64) = log .10\n\nn = (log .10)/(log .64)\n\nn = -1/-.1938\n\nn = 5.16\n\nTherefore, 6 reductions would be needed.\n\n***EDIT***\nmathgirl826, after 1 reduction, there is 64% left.
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...
1121982
look std 10 th NCERT mathematics text book... its in the part 2 of  mathematics in geometry section.
886843
(1)\nh = 3\nk = 4\n5 = a(0-3)² + 4\n5 = 9a + 4\n1 = 9a\n(1/9) = a\n\nY = (1/9)(x - 3)² + 4\n\n(3)\nY = (¼)x² + 4x -  7\nY = (¼)(x²+16x) - 7\n\n x² +16x = x² + 16x + (16/2)² - (16/2)²\n     = [x + (16/2)]² - (16/2)\n     = (x + 8)² - 8²\n     = (x + 8)² - 64\n\nY = (¼)[(x + 8)² - 64] - 7\nY = (¼)(x + 8)² - 16 - 7\nY = (¼)(x + 8)² - 23
487838
1) Assuming an aritmetic sequence\n\nbn = b1 + (n-1)d \nletting b3=a1 = 20, then b7=a5 = 32, and b25 = b23\na5 = a1 + (5-1)d\n32 = 20 + (5-1)d\n32 = 20 + 4d\n12 = 4d\n3 = d\n\nb25 = a23 = 20 + (23-1)*3\nb25 = 20 + 22*3\nb25 = 20 + 66\nb25 = 86\n=====================================================\n2)\na75 = a1 + (75-1)d; d=-3\n\na75 = 5 + (75-1)*-3\na75 = 5 + 74*-3\na75 = 5 + -222\na75 = -217\n\nSn = n/2(a1+an)\nS75 = 75/2(5 + -217)\nS75 = 75/2(-212)\nS75 = -7950\n===================================================\n3) ????\n\na5 = 5*5 - 4 = 21\na15 = 5*15 - 4 = 71\na25 = 5*25 - 4 = 121\n\nOR MAYBE\n\na1 = 5*1 - 4\na1 = 5 - 4\na1 = 1\na30 = 5*30 - 4\na30 = 150 - 4\na30 = 146\n\nSn = n/2(a1+an)\nS30 = 30/2(1 + 146)\nS30 = 30/2(147)\nS30 = 2205\n================================================\n4)\n\nrth term is nCr-1m^n-(r-1)k^r-1\n\n5th term =\n4C4m^4-(5-1)k^5-1\n4C4 = 1\n\n5th term = k^4
33162
Check Out This Website!! \n\nGives You Examples And Everything!! \n\nhttp://www.mathleague.com/help/geometry/geometry.htm
290472
This is a geometric sum. Consider the sum S = a^0 + a^1 + ... + a^n for some integer n. Then S - 1 = a^1 + ... + a^n and (S-1)/a = a^0 + a^1 + ... + a^(n-1). If we take the difference S - [(S-1)/a], we get a^n - a^(n-1). Solving for S, we get S = [1 - a^(n+1)] / [1 - a].\nThis theorem applies to your series for a=3 and n=7. You'll have to add the 1/3 in afterwards.
825795
OK,as the first one is already done,I am doing the rest for you\n4(2x-6)=x+2 =>8x-24=x+2  =>8x-x=2+24  =>7x=26  =>x=26/3=8 2/3  (3) 2(x+5)+1=3(x-1)+4   =>2x+10+1=3x-3+4 =>2x-3x=-3+4-10-1  => -x=-4  =>x=1  (5)x+6=4(4x-1)   =>x+6=16x-4  =>x-16x=-4-6  => -15x=-10  =>x=-10/-15=2/3 [after reducing]
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!!^_^
904643
This is a simple geometric sequence with the formula x^x (the little ^ sign means "to the power of")\n\nSo we have 1^1 (or 1*1) = 1\n2^2 (or 2*2) = 4\n3^3 (or 3*3*3) = 27\n4^4 (or 4*4*4) = 256\nso next is... 5^5\n5*5*5*5*5 = 3125
971807
(0,0)\n(3,0)\ndraw the graphs and then find them out!\nif u can't, then mail me!
656598
try seeing OEIS: http://www.research.att.com/~njas/sequences/?q=1%2C+1%2C+3%2C+5%2C+11%2C+21%2C+43&sort=0&fmt=0&language=english&go=Search\n\neverytime you had an integer sequence problem, you could check there, most of the time, they had an entry of the possible sequence there.\n\nthe most possible answer to your question is perhaps the \nJacobsthal sequence: a(n) = a(n-1) + 2a(n-2). \n\nwhich gives the sequence\n0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, 1365, 2731, 5461, 10923, 21845, 43691, 87381, 174763, 349525, 699051, 1398101, 2796203, 5592405, 11184811, 22369621, 44739243, 89478485, 178956971, 357913941, 715827883, 1431655765, 2863311531, ....
844539
1. An arithmetic sequence has a(sub 5)= -10 and a(sub 12) =18. Find a(sub 2) and a(sub 17)\n\nLet a(sub 1) = a and d = common difference.\nFor an AP:\na(sub n) = a + (n - 1)d\n\nSo a(sub 5) = a + 4d = -10 .... (1)\nand a(sub12) = a + 11d = 18 . (2)\nEquation (2) - equuation (1)\n7d = 28\nSo d = 4\nSo a = -26\na(sub 2) = a + d = -22\na(sub 17) = a + 16d = 38\n\n2. Find s(sub 4) for the geomtric sequence: \na(sub 1) = 5p, a(sub 2) = -10p\n\nLet a(sub 1) = a and common ratio be r\n\nFor a GP\na(sub n) = ar^(n-1)\n\na(sub 1) = a = 5p .......... (1)\na(sub 2) = ar =  -10p ..... (2)\nEquation (2)/equation (1)\nr = -2\n\nSo a(sub n) = 5p * (-2)^(n-1)\n\ns(sub n) = a(r^n - 1)/(r - 1)\n= 5p((-2)^n) - 1)/(-2 - 1)\n= -5p/3 ((-2)^n - 1)\n\nSo s(sub 4) = -5p/3 ((-2)^4 - 1)\n= -25p
664851
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.
1066144
Without the knowledge of where point e is, this problem is insolvable.\n\nI suspect, however, from a brief diagramming session, that you will need to solve for the length of ab using the Pythagorean Theorem, and set up similar triangles.
432855
r = a_{n+1}/a_n -> 1 if <a_n> is a sequence of Real values that converges to a non-zero L.\n\na_{n+1}/a_n = (L + (a_{n+1} - L))/(L + (a_n - L)).  But (a_{n+1} - L) -> 0 and (a_n - L) -> 0 since a_n -> L so that (L + (a_{n+1} - L))/(L + (a_n - L)) -> (L+0)/(L+0) = L/L = 1
847850
Geometric series: (termFirst - termLastPlusOne)/(1 - ratio):\n(e^x -e^(7x))/(1-e^x)