GATE CS 2014 Set 2 QGA-4 — General Aptitude

The multiples of $10$ lying between $2$ and $198$ are:

$$10, 20, 30, \dots, 190$$

These numbers form an Arithmetic Progression (AP) where:

• First term ($a$) = $10$
• Last term ($l$) = $190$
• Common difference ($d$) = $10$

For any Arithmetic Progression, the average of all the terms is equal to the average of the first and the last term:

$$\text{Average} = \frac{\text{First term} + \text{Last term}}{2}$$

$$\text{Average} = \frac{10 + 190}{2} = \frac{200}{2} = 100$$

Therefore, the correct option is B.