Question

Consider the sequence 2, 3, 5, 6, 7, 8, 10, 11, ..... of all positive integer, then 2011th term of this

sequence is​

Answer 1

Given : sequence 2, 3, 5, 6, 7, 8, 10, 11, ..... of all positive integer

To find : 2011th term of sequence

Solution:

2, 3, 5, 6, 7, 8, 10, 11,

Sequence is missing square terms

1² , 2² , 3² and so  on

1936  = 44²

upto 2011   there will me missing 44 terms    

Hence 2011 + 44  = 2055  terms should be there

2025 = 45²

but again  2025 < 2055  hence

45² is also missing there

Hence 2055 + 1 = 2056

46² = 2116 is greater than 2056

Hence  2011th term of this  sequence  is 2056

Learn more:

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Answer 2

Answer:

Since 45

2

=2025 and 46

2

=2116, there are precisely 45 perfect squares ≤2056 which are left out from the sequence of positive integers. Since 2056−45=2011, we conclude that the 2011

th

element is 2056