Now, what do you do in a sequence when you have nonconsecutive terms?
924311
1.)\nan = a1 * r^(n - 1)\n\na5 = a1 * r^(5 - 1)\n(1/24) = (2/3) * r^4\n(3/24) = 2r^4\n(1/8) = 2r^4\nr^4 = (1/16)\nr = (1/2)\n\nan = (2/3)(1/2)^(n - 1)\n\na1 = (2/3)(1/2)^(1 - 1) = (2/3)(1/2)^0 = (2/3) * 1 = (2/3)\na2 = (2/3)(1/2)^(2 - 1) = (2/3)(1/2)^1 = (2/3)(1/2) = (1/3)\na3 = (2/3)(1/2)^(3 - 1) = (2/3)(1/2)^2 = (2/3)(1/4) = (1/6)\na4 = (2/3)(1/2)^(4 - 1) = (2/3)(1/2)^3 = (2/3)(1/8) = (1/12)\na5 = (2/3)(1/2)^(5 - 1) = (2/3)(1/2)^4 = (2/3)(1/16) = (1/24)\na6 = (2/3)(1/2)^(6 - 1) = (2/3)(1/2)^5 = (2/3)(1/32) = (1/48)\n\na6 = (1/48)\n\n---------------------------------------------\n\n2.)\nan = a1 + (n - 1)d\n\na5 = a1 + (5 - 1)d\n-5 = a1 + 4d\na1 = -4d - 5\n\na7 = a1 + (7 - 1)d\n-2m - 11 = a1 + 6d\na1 = -2m - 11 - 6d\n\n-4d - 5 = -2m - 11 - 6d\n2d = -2m - 6\n2d = 2(-m - 3)\nd = -m - 3\n\na1 = -4(-m - 3) - 5 = 4m + 12 - 5 = 4m + 7\n\na1 = (4m + 7) + (1 - 1)(-m - 3)\na1 = 4m + 7 + 0(-m - 3)\na1 = 4m + 7\n\na2 = 4m + 7 + (2 - 1)(-m - 3)\na2 = 4m + 7 + 1(-m - 3)\na2 = 4m + 7 - m - 3\na2 = 3m + 4\n\na3 = 4m + 7 + (3 - 1)(-m - 3)\na3 = 4m + 7 + 2(-m - 3)\na3 = 4m + 7 - 2m - 6\na3 = 2m + 1\n\na4 = 4m + 7 + 3(-m - 3)\na4 = 4m + 7 - 3m - 9\na4 = m - 2\n\na5 = 4m + 7 + 4(-m - 3)\na5 = 4m + 7 - 4m - 12\na5 = -5\n\na6 = 4m + 7 + 5(-m - 3)\na6 = 4m + 7 - 5m - 15\na6 = -m - 8\n\na7 = 4m + 7 + 6(-m - 3)\na7 = 4m + 7 - 6m - 18\na7 = -2m - 11\n\na8 = 4m + 7 + 7(-m - 3)\na8 = 4m + 7 - 7m - 21\na8 = -3m - 14\n\nANS : a8 = -3m - 14
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.
297783
you start with one, than add zero.  Add one, add zero.  Add one, add zero.  Continue until you get tired.  Than repeat.
644170
The answer of "nayanmange" is right but his formula is wrong because it does not predict the sequence correctly. The correct formula is (2n-1)-th and 2n-th term of the series are {(2n-1)^2+1} and {(2n)^2-1} respectively. For n = 3 we find the (2*3)-th i.e. 6th term of the series as {(2*3)^2-1} = 35.
571951
Start at 2, and eliminate every second number after that.\nThen start at 3, and eliminate every third number after that.\n4 will already have been eliminated in step 1.\nStart at 5, and eliminate every fifth number after that.\nContinue this way.  The numbers left uneliminated are prime.
222592
1. p'q\n\n2. p'\n\n3. pq' + p'\n\nI think i got the correct ones !
247548
You can construct one simply by interleaving a bounded sequence with an unbounded sequence. For example, the sequence 1/n with the sequence n to form \n\nS(i) = if i is odd 1/i if i is even i. \n\nThis has 1/i as a convergent subsequence.
589983
Add some rigor. Remember that a sequence {a_i} is defind to converge to a limit L iff for every positive difference D > 0, there exists some n such that for all N > n, the difference between the limit L and any subsequent term a_N is less than D. We are squeezing D down to 0 by showing that we can always supply such an n. Thus, your job is to supply some pseudofunction n(D) giving an upper range of indices that satisfy the definition above.\n\nFor the pattern above, the (2n + 1)th term is 8 + 1/(n + 2) while the (2n)th term is 8 + 2/(n + 2). As n increases without bound, the (2n + 1) subsequence approaches 8, as given any positive difference D, one can find a least (2n + 1) such that all terms greater than (2n + 1) yields a difference 8 - (8 + 1/(n + 2)) = 1/(n + 2) less than D. In this case, choose n(D) > (1 - 2D)/D. \nUse a similar tactic to show that the even indexed subsequence converges to the same value, and thus the sequence converges to 8.\n\nAs for the second sequence, as long as all of the remaining values are 6, it converges to 6 by a trivial argument. Given a positive difference D, simply choose n(D) > 6, since 6 - a_n = 0 for any n > 6.
381290
I have an iterative method that is good for at least 80% of these types of problems.\n\nFirst, write down the series.\n\nNext, above the line you just wrote, write the difference between each pair of elements. In this case, it will be 1, 2, 4, 8, .... In this case, the answer is obvious - they are powers of two. The next one in the differences line should be 16, so the next on the original line should be 15+16=31. This will allow you to get any member in the sequence.\n\nIf the first order of differences doesn't give rise to a discernable pattern, try the differences of differences and so on.
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.
738502
According to the mnemonic 'PEMDAS' or 'Please Excuse My Dear Aunt Sally', the order is:\n\nParentheses\nExponents\nMultiplication\nDivision\nAddition\nSubtraction\n\nSo exponents come before multiplication
1125943
The order of operations are :\nParenthesis\nExponents\nMultiply\nDivide\nAdd\nSubtract
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]
823906
not quite sure what exactly you mean, but i would guess\n\n2+2+2+2 or 4+4 or 1+1+1+1+1+1+1+1
879402
Try:\n\nConfess\nAcknowledge\nOwn\nConcede
929018
mn=20!\nmn= 20*19*18*17*......*2=20!\nif m=20; n=19!\nif m=20*19; n=18!\nand so on.\nso m=20!/n; n=i! .... i 20 to 1\n\n0<m/n<1 means that m Not eq 0; and m < n\nby solving the above,\nm will remain < n only till iteration 8\n\nSo the ordered pairs are:\nm=20!/20!; n= 20!\nm=20!/19!; n=19!\nm=20!/18!; n=18!\nm=20!/17!; n=17!\nm=20!/16!; n=16!\nm=20!/15!; n=15!\nm=20!/14!; n=14!\nm=20!/13!; n=13!
942547
Can you please check you haven't made a mistake somewhere? I have just calculated (with 99% certainty) that there are no solutions to those equations with those restrictions.\n\nEdit - OK, I assumed you made exactly one mistake. By removing each possible word and trying to solve it, I discovered that removing 'SIGH' was the only option that gave an answer.\nt 1\nh 8\na 3\nn 7\nk 6\ns 5\ng 9\ni 4\nv 2\nwhere SIGH = 26, not 25. Then THANKSGIVING = 65.\n\nThat was actually a cooler puzzle that the original, assuming you made one mistake :)