Question

The common difference of the A.P.: $3,\,3+\sqrt{2},\,3+2\sqrt{2},\,3+3\sqrt{2},\,\ldots$ will be:
 

• A. $1+\sqrt{2}$
• B. $3(1+\sqrt{2})$
• C. $2\sqrt{2}$
• D. $\sqrt{2}$

Hint:

To find the common difference of an A.P., always subtract any two consecutive terms. The difference remains constant throughout.

Answer 1

The Correct Option is D

Step 1: Recall definition of common difference
In an arithmetic progression (A.P.), the common difference $d$ is given by: \[ d = a_{n+1} - a_n \]

Step 2: Take consecutive terms
First term = $3$, second term = $3+\sqrt{2}$.
\[ d = (3+\sqrt{2}) - 3 = \sqrt{2} \]

Step 3: Verify with next terms
Third term = $3+2\sqrt{2}$. Difference from second term: \[ (3+2\sqrt{2}) - (3+\sqrt{2}) = \sqrt{2} \] This matches. Hence $d = \sqrt{2}$.
\[ \boxed{d = \sqrt{2}} \]