what is the 2005th term of the sequences: 1,2,2,3,3,3,4,4,4,4,5,5,5,5,5?
101387
It will be 63. In fact, it will be 63 from the 1953rd position until the 2016th position, at which point it will be 64
750837
1--d\n2--a\n3--b\n4--a
754998
a 4\nb 5\nc 3\nd 2\ne 3\nf 4\ng 3\nh 5\ni 5\nj 5\nk 3\nl 5\nm 5\nn 5
291700
1\n2\n3\n4\n5\n6\n7\n8\n9\n10
465985
1 - D\n2 - B\n3 - C\n4 - C
21468
It depends on the sequence of course and the values. The use of the word "greater" here is very vague. The 50th term will always be 5 terms ahead of the 45th term, regardless of n. You wouldn't say it is 5 GREATER though. Sequences will have different values for the 50th and 45th term, so you're question cannot truly be answered without more information on the nature of the sequence.
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
1014029
1,2,3,5,16,63, ...\n3=(1+2)*1\n5=(2+3)*1\n16=(3+5)*2\n63=(5+16)*3
654672
Bingo!  I've got it!\n\nConvert from base-3 to base-10:\n3, 5, 8, 11, 16...\n\nSubtract n from each term:\n2, 3, 5, 7, 11...\n\nThis is the sequence of prime numbers!  So the generating sequence would be the nth prime plus n:\n2+1, 3+2, 5+3, 7+4, 11+5, 13+6, 17+7, ...\n\nContinuing the sequence:\n3, 5, 8, 11, 16, 19, 24, 27, 32, 39, 42, 49, 54, 57, 62, 69, 76, 79, 86, 91, 94, 101, 106, 113, 122, 127, 130, 135, 138, 143, 158, 163, 170, 173, 184, 187, 194, 201, 206, 213, 220, 223, 234, 237, 242, 245, 258, 271, 276, 279, 284, 291, 294, 305, 312, 319, 326...\n\nIn base-3:\n10, 12, 22, 102, 121, 201, 220, 1000, 1012, 1210, 2000, 2010, 2022, 2120, 2211, 2221, 10012, ...\n\nAm I right?
527322
1+1-21+30-5\n2-21+30-5\n-19+30-5\n11-5\n6
367516
here are the anwsers in order from top to bottom \n-T\n-T\n-F\n-F \nhope this helps
197550
1\n2\n3\n4\n5\npowerball 6
808688
the 1st three terms means n=1, n=2, n=3 so substitute\n\nThis gives, 3*1+2,3*2+2,3*3+2\n\n=5,8,11
723065
The sequence is the given series is *2 *3 *4 *5 *6and so on...so the answer is 720
580984
1. c\n2. c\n3. b\n4. a\nand 5 is a.
869678
1 and 1\n9 and 3\n25 and 5\n49 and 7\n81 and 9\n121 and 11\n...
1080683
1113213211\n\nAfter starting the sequence with 1, each term in the sequence consists of groups of two numbers based on the previous term - the first being the quantity and the second specifying which digit.