$I f t h e m e a n o f t h e s e t o f n u m b e r x_{1}, x_{2} . . . x_{n} i s ¯ x, t h e n t h e m e a n o f t h e n u m b e r x_{i} + 2 i, 1 \leq i \leq n i s :$

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Solution

The correct option is 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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If the mean of the set of number x1,x2...xn is ¯x, then the mean of the numberxi+2i, 1≤i≤ n is:

The correct option is B ¯x=∑ni=1xin⇒∑ni=1xi=n¯x∴ ∑ni(xi+2i)n=∑ni=1xi+2(1+2+...+n)n=n¯x+2n(n+1)2n=¯x+(n+1)

$I f t h e m e a n o f t h e s e t o f n u m b e r x_{1}, x_{2} . . . x_{n} i s ¯ x, t h e n t h e m e a n o f t h e n u m b e r x_{i} + 2 i, 1 \leq i \leq n i s :$

The correct option is B
$¯ x = \frac{\sum_{i = 1}^{n} x_{i}}{n} \Rightarrow \sum_{i = 1}^{n} x_{i} = n ¯ x ∴ \frac{\sum_{i}^{n} (x_{i} + 2 i)}{n} = \frac{\sum_{i = 1}^{n} x_{i} + 2 (1 + 2 + . . . + n)}{n} = \frac{n ¯ x + 2 \frac{n (n + 1)}{2}}{n} = ¯ x + (n + 1)$