Question

show that root 3 root 12 root 27 root 48 is an ap​

Answer 1

$ap = \sqrt{3} . \sqrt{12} .\sqrt{27}. \sqrt{48}$

we know that in an A.P.

$t_{2} - t_{1} = t_{3} - t_{2}$

$\sqrt{12} - \sqrt{3} = \sqrt{27} - \sqrt{12}$

Answer 2

AP = t2- t1= t3- t2

t1= root 3

t2= root 12

t3 = root 27

root 12 - root 3 = root 27 - root 12