Question
Is the given sequence: $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots$ form an $AP?$ If it forms an $AP$, then find the common difference $d$ and write the next three terms.

Answer

From the given information, we can have
${a_{2}-a_{1}=\sqrt{6}-\sqrt{3}}$
${a_{3}-a_{2}=\sqrt{9}-\sqrt{6}=3-\sqrt{6}}$
$a_{4}-a_{3}=\sqrt{12}-\sqrt{9}=2 \sqrt{3}-3$ $(since\sqrt{12}=\sqrt{2} \times 2 \times 3=2 \sqrt{3}) $
since $a_{k+1}-a_{k}$ is not the same for all values of $k$.
Hence, it is not an $AP$.

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