Q.Find the 6th term of the Geometric Progression $2, 6, 18, 54, \ldots$

Identify $a$ and $r$, then substitute into $t_n = ar^{n-1}$.

Here $a = 2$ and $r = \dfrac{6}{2} = 3$ (checked: $\dfrac{18}{6} = 3$, $\dfrac{54}{18} = 3$ — constant, so this is a GP).

Apply $t_n = a\,r^{n-1}$ with $n = 6$:

$t_6 = 2 \times 3^{6-1} = 2 \times 3^{5} = 2 \times 243 = 486.$ …

Split the power: $3^5 = 3^2 \times 3^3 = 9 \times 27 = 243$, then $\times 2 = 486$ — a way to …

Using the exponent $n$ instead of $(n-1)$: $2 \times 3^{6} = 1458$ is actually the 7th term. The exponent on …