If variance of first n natural number is 10 and variance of first m even natural number is 16, then the value of m+n is
Variance=n∑i=1x2in−(¯x)2
10=n(n+1)(2n+1)6n−[n(n+1)2n]2
=(n+1)(2n+1)6−(n+1)24
=(n+1)24[4(2n+1)−6(n+1)]
⇒(n+1)(n−1)=2402
⇒n2−1=120
⇒n=11
Similarly, first m even numbers are 2,4,6,⋯2m, we have
16=4(12+22+32+42+⋯+m2)m−[2(1+2+3+4+⋯⋯+m)m]2
=4m(m+1)(2m+1)6m−[2m(m+1)2m]2
⇒(m+1)[23(2m+1)−(m+1)]=16
⇒(m+1)(m−1)=48 ⇒m2−1=48
⇒m=7
∴m+n=7+11=18