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Question

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

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Solution

Variance=ni=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)(n1)=2402
n21=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)(m1)=48 m21=48
m=7
m+n=7+11=18


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