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(B)r(C)rp-q(D)r+[16JM11006Consider the sequence 2,3, 5, 6, 7, 8, 10, 11,. (A) 2056of all positive integer, then 2011t term of this sequence(C) 2013(B) 2011(D) 2060 |
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Answer» we can see that here 4,9 are missing... so, the sequence doesn't contain all the square numbers now √2011 = 44.84.. so nearest perfect square is of 45*45 = 2025....so there will be 45 such numbers missing im the sequence.. and so the sequence will go for 45 ahead of 2011 so the number is 2011+45 = 2056. option A. |
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