If the mean of the set of number x 1 , x 2 . . . x n is ¯ x ,then the mean of the number x i + 2 i , 1 ≤ ≤ n is Options: A . &nbsp¯x+2n B . &nbsp¯x+n+1 C . &nbsp¯x+2 D . &nbsp¯x+n

Answer: Option B : B ¯ x = ∑ n i = 1 x i n ⇒ ∑ n i = 1 x i = n ¯ x ∴ ∑ n i ( x i + 2 i ) n = ∑ n i = 1 x i + 2 ( 1 + 2 + . . . + n ) n = n ¯ x + 2 n ( n + 1 ) 2 n = ¯ x + ( n + 1 )

If the mean of the set of number x1,x2...xn is ¯x ,then the mean of t

Question

If the mean of the set of number

x_{1}, x_{2} . . . x_{n}

¯ x

,then the mean of the number

x_{i} + 2 i, 1 \leq\leq n

Options:

A . ¯x+2n

B . ¯x+n+1

C . ¯x+2

D . ¯x+n

Answer: Option B
:
B

¯ x = \frac{\sum_{i = 1}^{n} x_{i}}{n} \Rightarrow \sum_{i = 1}^{n} x_{i} = n ¯ x ∴ \frac{\sum_{i}^{n} (x_{i} + 2 i)}{n} = \frac{\sum_{i = 1}^{n} x_{i} + 2 (1 + 2 + . . . + n)}{n} = \frac{n ¯ x + 2 \frac{n (n + 1)}{2}}{n} = ¯ x + (n + 1)

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