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: Understanding the Question:
The topic of this question is Arithmetic Progressions (A.P.).
An Arithmetic Progression is a sequence of numbers in which the difference between any two consecutive terms is a constant.
This constant difference is called the common difference, typically denoted by \(d\).
We are given the general formula for the \(n^{\text{th}}\) term of an A.P., which is \(a_n = \sqrt{2}n + 1\).
Our objective is to calculate the common difference of this progression.

Step 2: Key Formula or Approach:
The common difference \(d\) of an A.P. whose general term is \(a_n\) can be calculated as the difference between any term and its preceding term:
\[ d = a_n - a_{n-1} \] Alternatively, we can find the common difference by calculating the first term (\(a_1\)) and the second term (\(a_2\)), and then finding their difference:
\[ d = a_2 - a_1 \] We will apply both methods to verify our calculation.

Step 3: Detailed Explanation:

• Let the \(n^{\text{th}}\) term of the sequence be represented by \(a_n = \sqrt{2}n + 1\).

• Find the first term of the A.P. (\(a_1\)) by substituting \(n = 1\) into the expression:
\[ a_1 = \sqrt{2}(1) + 1 = \sqrt{2} + 1 \]

• Find the second term of the A.P. (\(a_2\)) by substituting \(n = 2\) into the expression:
\[ a_2 = \sqrt{2}(2) + 1 = 2\sqrt{2} + 1 \]

• Compute the common difference \(d\) by subtracting \(a_1\) from \(a_2\):
\[ d = a_2 - a_1 \] \[ d = (2\sqrt{2} + 1) - (\sqrt{2} + 1) \] \[ d = 2\sqrt{2} + 1 - \sqrt{2} - 1 \] \[ d = \sqrt{2} \]

• Alternatively, we can verify this by calculating \(a_n - a_{n-1}\) in general terms:
\[ a_{n-1} = \sqrt{2}(n-1) + 1 = \sqrt{2}n - \sqrt{2} + 1 \] Now, subtract this from \(a_n\):
\[ a_n - a_{n-1} = (\sqrt{2}n + 1) - (\sqrt{2}n - \sqrt{2} + 1) \] \[ a_n - a_{n-1} = \sqrt{2}n + 1 - \sqrt{2}n + \sqrt{2} - 1 \] \[ a_n - a_{n-1} = \sqrt{2} \] This general algebraic verification confirms that the difference between any two consecutive terms is constant and equals \(\sqrt{2}\).


Step 4: Final Answer:
The common difference of the given Arithmetic Progression is \(\sqrt{2}\).
Therefore, the correct option is (A).