Question

The 6th term of the AP \( \sqrt{27}, \sqrt{75}, \sqrt{147}, \ldots \) is:

• A. \( \sqrt{243} \)
• B. \( \sqrt{363} \)
• C. \( \sqrt{300} \)
• D. \( \sqrt{507} \)

Hint:

Convert all square roots to a common radical form to identify arithmetic patterns easily. Use the nth term formula for AP once the common difference is clear.

Answer 1

The Correct Option is D

Step 1: Write the first three terms of the AP: \[ a_1 = \sqrt{27}, \quad a_2 = \sqrt{75}, \quad a_3 = \sqrt{147} \] Step 2: Simplify each square root in terms of \( \sqrt{3} \): \[ \sqrt{27} = \sqrt{3 \times 9} = 3\sqrt{3}, \\ \sqrt{75} = \sqrt{3 \times 25} = 5\sqrt{3}, \\ \sqrt{147} = \sqrt{3 \times 49} = 7\sqrt{3} \] Step 3: Now rewrite the sequence: \[ a_1 = 3\sqrt{3}, \quad a_2 = 5\sqrt{3}, \quad a_3 = 7\sqrt{3} \] Step 4: Find the common difference: \[ d = a_2 - a_1 = 5\sqrt{3} - 3\sqrt{3} = 2\sqrt{3} \] Step 5: Use the nth term formula of AP: \[ T_n = a + (n - 1)d \] Step 6: Put values to find 6th term: \[ T_6 = a + 5d = 3\sqrt{3} + 5 \cdot 2\sqrt{3} = 3\sqrt{3} + 10\sqrt{3} = 13\sqrt{3} \] Step 7: Convert back to square root form: \[ T_6 = 13\sqrt{3} = \sqrt{(13\sqrt{3})^2} = \sqrt{169 \cdot 3} = \sqrt{507} \] \[ \Rightarrow \boxed{T_6 = \sqrt{507}} \]