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

Identify $a$ and $r$, then substitute into the GP general-term formula.

$T_6=2\times3^{5}=2\times243=486$.

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

Apply $T_n=ar^{n-1}$. For $n=6$: $T_6=2\times3^{6-1}=2\times3^5=2\times243=486$.

Check by direct listing: $2,6,18,54,162,486$ — the $6$th term in this list is indeed $486$.

Alternative — repeated multiplication: starting from $a=2$, multiply by $3$ a total of $5$ times ($6-1$): $2\to6\to18\to54\to162\to486$, landing on $486$ at the $6$th term — the same result reached step by step.

As with an AP, students often use the exponent $n$ instead of $(n-1)$, giving the wrong value $2\times3^6=1458$ (which is actually the 7th term). The exponent on $r$ is always one less than the term's position.