101 is not a number sequence. So the question, as stated, makes no sense.
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To find the nth term of the sequence 11, 21, 35, 53, 75, 101, we can observe the differences between consecutive terms: 10, 14, 18, 22, and 26, which increase by 4 each time. This suggests that the sequence can be described by a quadratic function. The nth term can be represented as ( a_n = 5n^2 + 6n ), where n starts from 1. Thus, the nth term corresponds to this formula for values of n.
If the sequence is 205, 306, 427 then a possible 4th number is 568. The initial difference between 205 and 306 is 101, which increases by 20 at each step. 306 → 427 = 121 : 427 → 568 = 141
164,850 Calculated as follows: The lowest number is 101 and the highest number is 998. If you use the theory behind factorials, and pair the numbers, the lowest with the highest (101+998=1,099) then the next lowest and next highest (104+995=1,099), etc. they all add to 1,099. The sequence goes by threes: 101, 104, 107, etc There are 300 total numbers in the sequence (the first is 101, and 299 more (299x3=897)...897+101=998). The actual formula is {(998-101)/3} + 1 = 300 (the 1 is for the first number) So that makes 150 pairs of numbers that = 1,099. 150 x 1,099 = 164,850