Question

Verify the $AP:\sqrt{3},2\sqrt{3},3\sqrt{3},...$ and find the product of the next two terms:

Answer 1

We have
$2\sqrt{3}-\sqrt{3}=\sqrt{3}$
$3\sqrt{3}-2\sqrt{3}=\sqrt{3}$
This shows that difference of a term and the preceding term is always same.Hence, the given sequence form an AP.
Here first term $a=\sqrt{3}$ and common difference $d=\sqrt{3}$
Therefore,
$a_4=a+3d=\sqrt{3}+3\times \sqrt{3}$
$\Rightarrow a_4=4\sqrt{3}$
$a_5=a+4d=\sqrt{3}+4\times \sqrt{3}$
$\Rightarrow a_5=5\sqrt{3}$
$a_6=a+5d=\sqrt{3}+5\times \sqrt{3}$
$\Rightarrow a_6=6\sqrt{3}$
Hence, the next three terms are
$4\sqrt{3},5\sqrt{3},6\sqrt{3}$