If the standard deviation of the first n natural numbers is t then $$12t^{2} + 1 =$$

The standard deviation of n natural numbers = t 

Therefore, but we know that 

First n natural numbers are 1,2,3,…,n.

Mean=(1+2+3+…+n)÷n

=[n(n+1)/2]/n

(n+1)/2 …(i)

And, 1²+2²+3²+…+n²=n(n+1)(2n+1)/6

Or, (1²+2²+3²+…+n²)/n= {(n+1)(2n+1)/6} …(ii)

Variance of first n natural numbers $$\sqrt{t}$$  = {(1²+2²+3²+…+n²)/n} - (Mean)²

= {(n+1)(2n+1)/6}-{(n+1)²/4}

= (n+1)/12{4n+2-3(n+1)}

= (n+1)(n-1)/12

$$\sqrt{t}$$ = (n²-1)/12

and we have the equation given 

$$12t^{2} + 1 =$$ 

t^2 = -1\12  put the of t we get 

n^2 Answer