Given :
A sequence of numbers consists of natural numbers that are
not perfect squares.
To Find :
The 1000th member of this sequence .
Solution :
Now , [tex]31^2=961[/tex] and [tex]32^2=1024[/tex] .
So , there are 31 numbers before 1000 which are perfect square .
Therefore , the [tex]999^{th}[/tex] is 1000+31 = 1031 .
Also , the first perfect square after 1000 is 32 .
So , we have to continue to 1031 + 1 = 1032 to compensate for the 32nd square.
Therefore , the 1000th member of this sequence is 1032 .
Hence , this is the required solution .
