If the mean of the data x₁, x₂, x₃, ….., x_n is 'a', then the mean of the data x₁ + a, x₂ + a, …., x_n + a is :

1. a

2. 2a

3. $\dfrac{1}{2}$a

4. $\dfrac{a}{3}$

Mean of x₁, x₂, x₃, ….., x_n is 'a'

So,

$\dfrac{x$1 + x2 + \dots + x_n}{n} = a

x₁ + x₂ + ….. + x_n = na

∴ $\sum x_i = na$

New observation is :

x₁ + a, x₂ + a, …., x_n + a

$\Rightarrow \text{New Mean} = \dfrac{(x$1 + a) + (x2 + a) + \dots + (xn + a)}{n} \\[1em] \Rightarrow \text{New Mean} = \dfrac{(x1 + x2 + \dots + xn) + (a + a + \dots + n \text{ times})}{n} \\[1em] \Rightarrow \text{New Mean} = \dfrac{\sum x_i + na}{n}

Since, $\sum x_i = na$

∴ $\text{New Mean} = \dfrac{na + na}{n} = \dfrac{2na}{n}$ = 2a.

Hence, option 2 is the correct option.