Question

Find the next term of the
$\sqrt{3 } \: \: \sqrt{12} \: \: \sqrt{27} \: . ....$
​

Answer 1

First term (a) = t1=

$\sqrt{3}$

Second term=t2=

$\sqrt{12}$

third term=t3=

$\sqrt{27}$

Here d=t2-t1

$\sqrt{12 } - \sqrt{ 3} \\ \sqrt{4 \times 3} - \sqrt{3} \\ 2\sqrt{3} - \sqrt{3} \\ \sqrt{3}$

also d=t3-t2

$\sqrt{27} - \sqrt{12} \\ \sqrt{9 \times 3} - \sqrt{4 \times 3} \\ 3\sqrt{ 3} - 2 \sqrt{3} \\ \sqrt{3}$

There d=√3

Let next term be t4

Formula for nth term-

$tn = a + (n - 1)d$

Therefore modifying

Here n=4

$t4 = \sqrt{3} + (4 - 1) \sqrt{3}$

$t4 = \sqrt{3 } +(3) \sqrt{3}$

$t4 = \sqrt{3} + 3 \sqrt{3}$

$t4 = 4 \sqrt{3}$

I hope this helps