Question

If variance of first n natural numbers is 10 and variance of first m even natural numbers is 16 then the value of m + n is ?

[JEE MAINS - 2020]

Answer 1

26

Explanation:

m=10,n=16

value= m+n

= 10+16

= 26

Answer...

Answer 2

EXPLANATION.

Variance of first n natural number = 10.

Variance of first m even natural number = 16.

As we know that,

Formula of variance :

If x₁, x₂, x₃ ,,,,,xₙ = $\sf \dfrac{\displaystyle\sum\limits_{1}^{n} (x_i)^{2} }{n} - \mu^{2}$  μ is the mean value of the terms.

n natural numbers = 10.

1, 2, 3, 4, ,,,,,,,,,n = 10.

Put the value in equation, we get.

$\sf \implies \dfrac{1^{2} + 2^{2} + 3^{2}+ 4^{2} + .. n^{2} }{n} - \bigg[\dfrac{n(n + 1)}{2n} \bigg]^{2} = 10$

$\sf \implies \dfrac{1^{2} + 2^{2} + 3^{2}+ 4^{2} + .. n^{2} }{n} - \bigg[\dfrac{(n + 1)}{2} \bigg]^{2} = 10$

As we know that,

1² + 2² + 3² + ,,,,,n² = n(n + 1)(2n + 1)/6.

Put the values in the equation, we get.

$\sf \implies \dfrac{n(n + 1)(2n + 1) }{6n} - \bigg[\dfrac{(n + 1)}{2} \bigg]^{2} = 10$

$\sf \implies \dfrac{(n + 1)(2n + 1) }{6} - \bigg[\dfrac{(n + 1)}{2} \bigg]^{2} = 10$

$\sf \implies \dfrac{(n + 1)(2n + 1) }{6} - \dfrac{(n + 1)^{2} }{4} = 10$

$\sf \implies \dfrac{2n^{2}+ n + 2n + 1 }{6} - \dfrac{n^{2} + 1 + 2n}{4} = 10.$

$\sf \implies \dfrac{2(2n^{2} + 3n + 1) \ - 3(n^{2} + 1 + 2n)}{12} = 10.$

$\sf \implies \dfrac{4n^{2} + 6n + 2 - (3n^{2} + 3 + 6n)}{12} = 10.$

$\sf \implies \dfrac{4n^{2} +6n + 2 - 3n^{2} - 3 - 6n}{12} = 10.$

$\sf \implies \dfrac{n^{2} - 1}{12} = 10.$

$\sf \implies n^{2} - 1 = 120.$

$\sf \implies n^{2} = 121.$

$\sf \implies n = 11.$

Variance of first m even natural number = 16.

2 + 4 + 6 + 8 + ,,,,, 2m = 16.

$\sf \implies \dfrac{2^{2} + 4^{2} + 6^{2}+,,,,+2m^{2} }{m} - \bigg[\dfrac{2m(m + 1)}{2m}\bigg]^{2} = 16.$

$\sf \implies \dfrac{2^{2} + 4^{2} + 6^{2}+,,,,2m^{2} }{m} - (m + 1)^{2} = 16.$

$\sf \implies \dfrac{2^{2}(1^{2}+ 2^{2} + 3^{2} + ,,,,m^{2}) }{m} - (m + 1)^{2} = 16.$

As we know that,

Formula of :

1² + 2² + 3² + ,,,,m² = m(m + 1)(2m + 1)/6.

Put the formula in equation, we get.

$\sf \implies \dfrac{4m(m + 1)(2m + 1)}{6m} - (m + 1)^{2} = 16.$

$\sf \implies \dfrac{4(m + 1)(2m + 1)}{6} - (m + 1)^{2} = 16.$

$\sf \implies \dfrac{4(m + 1)(2m + 1)\ - 6(m + 1)^{2} }{6} = 16.$

$\sf \implies \dfrac{4[2m^{2}+ m + 2m + 1] - 6[m^{2} + 1 + 2m]}{6} = 16.$

$\sf \implies \dfrac{8m^{2} + 12m + 4 - 6m^{2} - 6 - 12m}{6} = 16.$

$\sf \implies \dfrac{2m^{2} - 2}{6} = 16.$

$\sf \implies 2m^{2} - 2 = 96.$

$\sf \implies 2m^{2} = 98.$

$\sf \implies m^{2} = 49.$

$\sf \implies m = 7.$

To find value of m + n.

⇒ n = 11.

⇒ m = 7.

⇒ m + n = 11 + 7 = 18.

Value of [m + n] is = 18.