Solve this

Given: GP is 2, 6, 18, 54….

The given GP is of the form, $a, a r, a r^{2}, a r^{3} \ldots$

Where r is the common ratio.

First term in the given GP, $a_{1}=a=2$

Second term in GP, $a_{2}=6$

Now, the common ratio, $r=\frac{a_{2}}{a_{1}}$

$r=\frac{6}{2}=3$

 Now, $\mathrm{n}^{\text {th }}$ term of GP is, $a_{n}=a r^{n-1}$

So, the $6^{\text {th }}$ term in the GP,

$\mathrm{a}_{6}=\mathrm{ar}^{5}$

$=2 \times 3^{5}$

$=486$

$\mathrm{n}^{\text {th }}$ term in the GP,

$a_{n}=a r^{n-1}$

$=2.3^{n-1}$

Hence, $6^{\text {th }}$ term $=486$ and $n^{\text {th }}$ term $=2.3^{n-1}$