Question

$a^2,b^2,c^2$ are 3 consecutive perfect squares. How many natural numbers are lying in between $a^{2}and{c}^2$, if $a>0$?

• A. a + b
• B. a + b + 1
• C. 2 (a + b) + 2
• D. 2 (a + b) + 1

Answer 1

The correct option is D

2 (a + b) + 1

Difference between two consecutive perfect squares,
$n^{2}and(n+1{)}^2$
$(n+1{)}^2-n^2=n^2+2n+1-n^2=2n+1$
If the difference between two natural numbers is $2n+1$, then there are $(2n+1)-1=2n$ numbers between them.
So, we find that between $n^{2}and(n+1{)}^2$ there are $2n$ numbers which are 1 less than the difference of two squares.
For example, between $3^{2}and{4}^2$ there are $6$ natural numbers.
Which is given by $(4^2-3^2)-1=2\times 3=6$
So, Number of natural numbers in between $a^{2}and{b}^2=2a$
Number of natural numbers in between $b^{2}and{c}^2=2b$

So total number of natural numbers in between $a^{2}and{c}^2=2a+2b+1$, (Since $b^2$ lies in between $a^{2}and{c}^2)$