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:

Think of terms like $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}$ as $1x, 2x, 3x$ where $x = \sqrt{2}$.
The step-to-step difference is simply $x$, which is $\sqrt{2}$.
This simplification makes calculations with surds much easier to visualize mentally.

Answer 1

The Correct Option is A

Step 1: Use the nth term formula instead of subtracting neighbouring terms.
Instead of finding $d$ from the first two terms, we can pull it out using the relation between any two terms of an AP, which is often faster when the terms already look complicated.

Step 2: Recall the relevant formula.
For an AP with first term $a$ and common difference $d$, the $n$th term is $a_n = a + (n-1)d$. So for any two terms $a_m$ and $a_n$:
\[ a_n - a_m = (n - m)d \implies d = \frac{a_n - a_m}{n - m} \]

Step 3: Apply it using the first and fourth terms.
Here $a_1 = \sqrt{2}$ and $a_4 = 4\sqrt{2}$. Using $n = 4$ and $m = 1$:
\[ d = \frac{a_4 - a_1}{4 - 1} = \frac{4\sqrt{2} - \sqrt{2}}{3} \]
Factor out $\sqrt{2}$ in the numerator:
\[ d = \frac{\sqrt{2}(4 - 1)}{3} = \frac{3\sqrt{2}}{3} \]
Cancel the 3:
\[ d = \sqrt{2} \]

Step 4: Check with a different pair of terms.
Using the second and third terms instead, $n=3$, $m=2$:
\[ d = \frac{3\sqrt{2} - 2\sqrt{2}}{3 - 2} = \frac{\sqrt{2}}{1} = \sqrt{2} \]
Same value, so the sequence is consistent as an AP with this common difference.

Step 5: Final answer.
The common difference is $\sqrt{2}$, which is option (A).
\[ \boxed{d = \sqrt{2}} \]