Question

FIND THE NEXT TERM OF AN AP
√27,√48.√75........

TAKE IT EASY AND SOLVE IT...........
PLS ITS URGENT................

Answer 1

√27 = 3√3
√48 = 4√3
√75 = 5√3 .

Common difference = 4√3 - 3√3 = √3 .

Next term = 5√3 + √3 = 6√3 = √108 .

Therefore, next term is √108

Answer 2

$The \ next \ term \ of \ an \ AP \ \sqrt{27}, \ \sqrt{48}, \ \sqrt{75}... \ is \ \sqrt{108}$

Given:

Three consecutive terms:

$\sqrt{27}, \sqrt{48}, \sqrt{75}$

Solution:

$First \ term = a = \sqrt{27}=\sqrt{3 \times 3 \times 3}=3 \sqrt{3}$

$Second \ term = \sqrt{48}=\sqrt{4 \times 4 \times 3}=4 \sqrt{3}$

$Third \ term = \sqrt{75}=\sqrt{5 \times 5 \times 3}=5 \sqrt{3}$

Common difference (d) = (Second term) - (First term)

$d=(4 \sqrt{3}-3 \sqrt{3})$

$=(5 \sqrt{3}-4 \sqrt{3})$

$d=\sqrt{3}$

So, the fourth term: (n=4)

$\Rightarrow(a+(n-1) d)$

$\Rightarrow(3 \sqrt{3}+(4-1) \sqrt{3}$

$\Rightarrow(3 \sqrt{3}+3 \sqrt{3})$

$\Rightarrow 6 \sqrt{3}$

$\Rightarrow \sqrt{6 \times 6 \times 3}$

$\Rightarrow \text { Fourth term }=\sqrt{108}$