Published by:
CGP EDU Academic Team
Published on: August 13, 2026
Consider the sequence 2, 3, 5, 6, 7, 8, 10, 11, ..... of all positive integer, then 2011 th term of this sequence is
Text Solution
Verified by ExpertsThe correct answer is:
A
45 2 = 2025 & 46 2 = 2116 ⇒ there are 45 squares ≤ 2056
which are left only from sequence of possible integers since 2056 = 2011 + 45
∴ 2011 th term = 2056
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