The 10th term of the sequence (\sqrt3,\sqrt{12}, \sqrt{27}..... | Mathematics | SolveeIt

Question

The 10th term of the sequence $\sqrt3,\sqrt{12}, \sqrt{27}.........$ is

$\sqrt{243}$

$\sqrt{300}$

$\sqrt{363}$

$\sqrt{432}$

Answer

$\sqrt{300}$

Explanation

Solution

$\sqrt{3},\sqrt{12}, \sqrt{27},.....$ can be written as $\sqrt{3.1}, \sqrt{3.4}, \sqrt{3.9}, .....$ i.e., $\sqrt{3.1^{2}} , \sqrt{3.2^{2}}, \sqrt{3.3^{2}} , .....$ $\therefore 10$ th term $= \sqrt{3.10^{2}}$ $= \sqrt{300}$

Mathematics Question on Sum of First n Terms of an AP

Solution