Which of the following form an AP? Justify your answer.
$ \sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \ldots $
Given:
Given sequence is \( \sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \ldots \)
To do:
We have to check whether the given sequence is an AP.
Solution:
In the given sequence,
$a_1=\sqrt3, a_2=\sqrt{12}=\sqrt{4\times3}=2\sqrt3, a_3=\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}, a_4=\sqrt{48}=\sqrt{16\times3}=4\sqrt{3}$
$a_2-a_1=2\sqrt3-\sqrt3=\sqrt3$
$a_3-a_2=3\sqrt3-2\sqrt3=\sqrt3$
$a_4-a_3=4\sqrt3-3\sqrt3=\sqrt3$
Here,
$a_2 - a_1 = a_3 - a_2=a_4 - a_3$
Therefore, the given sequence is an AP.
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