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Arithmetic Progressions — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsArithmetic Progressions2 Marks
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.
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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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