find the first term of the sequences 5,10,20,40,....which exceeds 10 000.?
143755
The first term that exceeds 1000 is the 12th term and its value is 10240
383876
In an arithmetic sequence, the formula for the nth term is\n\ninitial term + (common difference times (n-1))\n\nIn other words, you start with a number and keep adding the same number to that original number. In the sequence you gave (the odd numbers) the first number is 1 and you keeping adding 2. (The reason why the nth term is 1 + (n-1)2 is that you started adding 2 with the second term.)\n\nso the 101st term of that sequence would be\n\n1 + (100)2 = 201\n\nThe sum of the sequence, also known as an arithmetic series, has its  own formula:\n\nSum = (initial term + last term)*n/2\n\nThis comes out of the fact that if you add the first and last term, the result will be the same as adding the second and second-to-last term, which will be the same as adding the third and the third-to-last term, etc.\n\nSo for 20 terms of the sequence above, the first term is 1, the 20th term is 1+(2)19=39 and the sum of the sequence is (1+39)*20/2=400.\n\nI'll leave it up to you to do the sum of the first 30 terms.\n\nHope that helps.
771405
40.\nhttp://www.research.att.com/~njas/sequences/b000602.txt
1029214
First term = x\nSecond term = 2x\nthird = 2(2x) = 4x\nfourth = 8x\nfifth = 16x\nsixth = 32x\nseventh = 64x\neighth = 128x\n\n8th/5th = 128x/16x = 128/16 = 8/1
1089657
1)Linear Sequences : ex. a=2n , n>=1:\n1,2,4,8,16,32,...\n\n2)Quadratic Sequences : ex. a=n^2 , n>=2:\n4,9,16,25,36,49,64,81,....\n\n3)Cubic Sequences : ex. a=n^3 , n>=2:\n8,27,125,.....\n\n4)Polynomial Sequences : ex. a=2^n , n>=0:\n1,2,4,8,16,32,64,128,....\n\n5)Factorial Sequences : ex. a=n! , n>=0:\n1,1,2,6,24,120,....\n\nAll the above are devided into 2 major types : finite and infinite sequences...
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/
1157976
0000, 0001, 0002, 0003, 0004, 0005, 0006, 0007, 0008, 0009, 0010, 0011... etc, etc  untill u get to 9998, 9999\n\ntry it out, might take a long time but one combination will work :))
875039
The first five are simple\nP(5,1) + P(5,2) + P(5,3) + P(5,4) + P(5,5) \n5 + 20 + 60 + 120 + 120 = 325\n\nSix and seven are a little more complicated\n6!/2! + 7!/2!2!\n360 + 1260 = 1620\n\nEight is special because you limit the maximum.\n\nList of bad values\n9x,xxx,xxx\n7x,xxx,xxx\n69,xxx,xxx\n67,xxx,xxx\n66,xxx,xxx\n65,9xx,xxx\n65,7xx,xxx\n65,6xx,xxx\n\n8!/3!2! - (2 * 7!/3!2!) - (3 * 6!/3!) - (3 * 5!/3!)\n3360 - 840 - 360 - 60 = 2100\n\nNow add the results.\n325 + 1620 + 2100 = 4045
725478
Let's figure out the row first...\n1st row has the first number\n2nd row has 2nd and 3rd number\nnth row ends with the n(n+1)/2 th number.\n44th row ends with 990th number\n45th row ends with 1035th number.\n\nSo the 1000th number is the 10th number in the 45th row.  Now the a-th number in the b-th row is C(b-1, a-1).  Therefore our number is C(44,9) = 44!/(9!*35!) = 708,930,508.
723065
The sequence is the given series is *2 *3 *4 *5 *6and so on...so the answer is 720
724609
1000+40+1000+30+1000+20+1000\n+10+1000+0=5100\n\nor was there something else?
283337
FORTY\n+ TEN\n+ TEN\n--------\nSIXTY\n\n\n29786\n+ 850\n+ 850\n--------\n31486
49819
Hold on....10....20......25.....29..30....25..29\n\n25
872421
1. the number before 10 is 9\nmultiply by two\n9 *2 = 18\nsubtract 1\n18 - 1 = 17\n\nthen for 17\n16*2 = 32\n32 - 1 = 31\n\nthen for 31\n30 * 2 = 60\n60-1 = 58\n\nthen for 58\n58 * 2 = 116\n116-1=115\n\nthat is the first five numbers (including 10):\n10, 17, 61, 58, 116\n\n2. multiply two by three\n2 * 3 = 6\nadd one\n6 + 1 = 7\n\nthen for 7\n7*3 = 21\n21+1 = 22\n\nthen for 22\n22*3 = 66\n66+1=67\n\nthen for 67\n67*3 = 201\n201+1=202\n the first 5 numbers are 2, 7, 22, 67, 202
250641
ok that's A, B, C ....\nlet's have A as thew first term and the first term of first arithmetic progression\nB will be the start of second\nthat's A,C,E,G\nok that would be the first arithmetic progression we would not midn it because it is just upto 7\nwe would mind the second cause it's upto 8\nB,D,F,H\nthat's H\nsince B= 1716, and the common difference or d= 408\nso H is the fourth term of thios progression\nH= B+3d\n1716+ 3(408)= 1716+1224= 2940. sorry i'm not that sure. us hgould have one more term given to be sure.
129538
The next number is 30651.\nIt has a recursion, a_n=a_(n-1)^2+2*a_(n-2).
178354
if there are finitely many digits after the decimal point, count them, if there's n of them, write what you had behind the decimal point over 10^n, that is a 1 with n zeroes.\nIf there are infinitely many, find the moment when the digits begin repeating.  put the repeating sequence over as many nines as there are repeating digits.For a complex case, see the example (when there's a repeating and not repeating part.\nExample\n0.34356356356...=0.34+0.00356356356.... [notation is sometimes 0.00(356)]\n0.34=34/100=17/50\n0.00(356)=1/1000*0.(356)=1/1000*(356/999)=356/999000