Question

If the arithmetic mean of the numbers $x_{1},x_{2},x_{3},......,x_{n}$ is $\bar{x}$, then the arithmetic mean of numbers

$ax_{1} + b,ax_{2} + b,ax_{3} + b,........,ax_{n} + b$, where a, b are two constants would be

• A. $\bar{x}$
• B. $na\bar{x} + nb$
• C. $a\bar{x}$
• D. $a\bar{x} + b$

Answer 1

Answer: (D)

$a\bar{x} + b$

Answer 2

Required mean $= \frac{(ax_{1} + b) + (ax_{2} + b) + ..... + (ax_{n} + b)}{n}$

$= \frac{a(x_{1} + x_{2} + ..... + x_{n}) + nb}{n} = a\bar{x} + b,$

$\left( \because\mspace{6mu}\mspace{6mu}\mspace{6mu}\frac{x_{1} + x_{2} + ..... + x_{n}}{n} = \bar{x} \right)$.