Question

The common difference of the AP : $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots$ is :

• A. $\sqrt{2}$
• B. 1
• C. $2\sqrt{2}$
• D. $-\sqrt{2}$

Hint:

To avoid calculation errors with irrational terms, think of $\sqrt{2}$ as an algebraic variable, say $x$.
The sequence then becomes $x, 2x, 3x, 4x, \dots$
The common difference is simply $2x - x = x$, which is $\sqrt{2}$.

Answer 1

The Correct Option is A

Step 1: Use the general nth-term relation instead of adjacent terms.
For any AP, $a_n = a_1+(n-1)d$, so the gap between the 4th and 1st term is exactly $3d$, that is, $a_4-a_1=3d$.
Step 2: Plug in the given terms.
Here $a_1=\sqrt2$ and $a_4=4\sqrt2$, so $a_4-a_1 = 4\sqrt2-\sqrt2 = 3\sqrt2$.
Step 3: Solve for d.
Since $3d=3\sqrt2$, dividing both sides by 3 gives $d=\sqrt2$.
\[ \boxed{\sqrt{2}} \]