Question

A sequence of numbers consists of natural numbers that are

Answer 1

Given :

A sequence of numbers consists of natural numbers that are

not perfect squares.

To Find :

The 1000th member of this  sequence .

Solution :

Now , $31^2=961$ and $32^2=1024$ .

So , there are 31 numbers before 1000 which are perfect square .

Therefore , the $999^{th}$ 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 .