Question

10th term of the sequence $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, \dots$ will be

• A. $\sqrt{242}$
• B. $\sqrt{288}$
• C. $\sqrt{200}$
• D. $\sqrt{162}$

Hint:

To find any term in an A.P., use $a_n = a_1 + (n - 1)d$. If the terms involve radicals like $\sqrt{2}$, treat them as coefficients for arithmetic progression.

Answer 1

The Correct Option is B

Step 1: Identify the sequence.
Given sequence: $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, \dots$
This is an arithmetic progression (A.P.) where
First term $a_1 = \sqrt{2}$ and common difference $d = \sqrt{2}$.
Step 2: Find the 10th term.
Formula for nth term of A.P.:
\[ a_n = a_1 + (n - 1)d \] Substitute $a_1 = \sqrt{2}$, $d = \sqrt{2}$, $n = 10$:
\[ a_{10} = \sqrt{2} + (10 - 1)\sqrt{2} = 10\sqrt{2} \] Step 3: Express in square root form.
\[ 10\sqrt{2} = \sqrt{100 \times 2} = \sqrt{200} \] Correct option is (C) $\sqrt{200}$.