The sum of first `n` odd natural numbers is `2n-1` (b) `2n+1` (c) `n^2` (d) `n^2-1`

The mean of first n odd natural number is (a) (n+1)/2 (b) n/2 (c) n (d) n^2

The mean of first n odd natural numbers is (n^2)/(81) , then n= (a) 9 (b) 81 (c) 27 (d) 18

If the sum of first n even natural numbers is equal to k times the sum of first n odd natural number then k= 1/n b. (n-1)/n c. (n+1)/(2n) d. (n+1)/n

If the sum of first n even natural numbers is equal to k times the sum of first n odd natural number then k= a. 1/n b. (n-1)/n c. (n+1)/(2n) d. (n+1)/n

Show that mean and variance of first n natural numbers is (n+1)/(2) and (n^(2) - 1)/(12)

Statement - I : The variance of first n even natural numbers is (n^(2) - 1)/(4) Statement - II : The sum of first n natural numbers is (n(n+1))/(2) and the sum of the squares of first n natural numbers is (n(n+1)(2n+1))/(6)

Statement 1 The difference between the sum of the first 100 even natural numbers and the sum of the first 100 odd natural numbers is 100. Statement 2 The difference between the sum opf the first n even natural numbers and sum of the first n odd natural numbers is n.

If the n^(t h) term of an A.P. is 2n+1 , then the sum of first n terms of the A.P. is n(n-2) (b) n(n+2) (c) n(n+1) (d) n(n-1)

For some natural number 'n', the sum of the first 'n' natural numbers is 240 less than the sum of the first (n+5) natural numbers. Then n itself is the sum of how many natural numbers starting with 1.

Let S_(n) denote the sum of the cubes of the first n natural numbers and s_(n) denote the sum of first n natural numbers then (S_(n))/(s_(n)) is equal to (i) (n(n+1)(n+2))/6 (ii) (n( n+1))/2 (iii) (n^(2)+3n+2)/2 (iv) (2n+1)/3