Q.Which term of the Arithmetic Progression $5, 11, 17, \ldots$ is equal to $65$?

Set the general-term formula equal to $65$ and solve for $n$.

$(n-1)=\dfrac{60}{6}=10 \Rightarrow n=11$.

Here $a=5$, $d=11-5=6$ (checked: $17-11=6$ — constant).

Set $T_n=65$: $5+(n-1)(6)=65 \Rightarrow (n-1)(6)=60 \Rightarrow n-1=10 \Rightarrow n=11$.

Since $n=11$ is a positive integer, $65$ genuinely is a term of this AP, specifically the $11$th.

Check by direct listing: $5,11,17,23,29,35,41,47,53,59,65$ — the $11$th number in this list is indeed $65$.

Alternative — express $65$ as "$5$ plus some whole number of steps of $6$": $65-5=60$, and $60/6=10$ whole steps, so $65$ is $10$ steps after the first term, i.e. the $(10+1)=11$th term — the same reasoning without formally writing the formula.

After solving $(n-1)=10$, a common slip is reporting the answer as $n=10$ (forgetting the final $+1$ step) instead of $n=11$. Always solve fully for $n$ itself, not just for $(n-1)$.