Question

find the 10th term of an ap root(3),root(12), root(27),..............

Answer 1

d=b-a=√12-√3
a=√3
tn=a+(n-1)d
t10=√3+9(√12-√3)
=√3+9√12-9√3
=9√12-8√3
=10√3

Answer 2

$\underline{\bold{Answer:-}}$

Given AP:-

$\sqrt{3} , \: \sqrt{12} , \: \sqrt{27} ,.........$


It can be written as :-

$\sqrt{3} , \: 2 \sqrt{3} , \:3 \sqrt{3} ,.......$

First term a = √3


Common Difference

$d = 2 \sqrt{3} - \sqrt{3} \\ \\ = > d = \sqrt{3} (2 - 1) \\ \\ = > d = \sqrt{3}$

We know that Tn is given by :-

Tn = a + ( n - 1 )*d



T10 = a + ( 10 - 1 )* d


=> T10 = a + 9d


=> T10 = √3 + 9 ( √3 )


=> T10 = √3 + 9√3


=> T10 = √3 ( 1 + 9 )


=> T10 = 10√3.



$\textbf{Hope it helps!}$