Question

The \(n^{\text{th}}\) term of an A.P. is \(\sqrt{2}n + 1\). Its common difference is

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

Hint:

For any Arithmetic Progression where the \(n^{\text{th}}\) term is expressed as a linear function of \(n\) of the form \(a_n = An + B\), the common difference \(d\) is always equal to the coefficient of \(n\), which is \(A\).
Here, the coefficient of \(n\) in \(\sqrt{2}n + 1\) is \(\sqrt{2}\), so you can write down the answer directly without any calculations!

Answer 1

The Correct Option is A

Step 1: Recognize the pattern of the nth term.
The nth term is \(a_n = \sqrt{2}n + 1\), which is a linear expression in n of the form \(An + B\).
Step 2: Recall what this form means for an AP.
For any AP written as \(a_n = An + B\), the coefficient of n is always equal to the common difference, because each time n increases by 1, the term increases by exactly A.
Step 3: Confirm using two terms further along the sequence.
Take \(a_5 = \sqrt{2}(5) + 1 = 5\sqrt{2} + 1\) and \(a_6 = \sqrt{2}(6) + 1 = 6\sqrt{2} + 1\). \(d = a_6 - a_5 = (6\sqrt{2}+1) - (5\sqrt{2}+1) = \sqrt{2}\), confirming the pattern from Step 2.
Step 4: State the final answer.
The common difference is \(\sqrt{2}\), matching option (A).
\[ \boxed{d = \sqrt{2}} \]