NIMCET 2018 — Mathematics PYQ

Concept:

Average/Mean Mean of 'n' observations =  
$$\bar{x}=\frac{\text{Sum of Observations}}{n}=\frac{{\sum }_{i=1}^n{x}_i}{n}$$  

Sum of the squares of the first n natural numbers =  
$$\sum n^2=\frac{n(n+1)(2n+1)}{6}$$  

Calculation:

We know that the sum of the squares of the first n natural numbers is: 
$$\frac{n(n+1)(2n+1)}{6}$$  

Therefore, Mean of the squares of the first n natural numbers is:  
$$\frac{\frac{n(n+1)(2n+1)}{6}}{n}=\frac{(n+1)(2n+1)}{6}$$  

According to the question: Mean = 11.

$$\frac{(n+1)(2n+1)}{6}=11$$  

$$2n^2+n+2n+1=66$$  

$$2n^2+3n-65=0$$  

$$2n^2+13n-10n-65=0$$  

$$n(2n+13)-5(2n+13)=0$$  

$$(2n+13)(n-5)=0$$  

$$2n+13=0\text{OR}n-5=0$$  

$$n=-\frac{13}{2}\text{OR}n=5$$  

Since n is a count, n = 5. 

Sum of the squares of the first 5 natural numbers =  
$$1^2+2^2+3^2+4^2+5^2=1+4+9+16+25=55$$  

Mean = $$\frac{55}{5}=11$$