Q.Which term of the Arithmetic Progression $5, 11, 17, 23, \ldots$ is equal to $95$?

Set the nth-term formula equal to $95$ and solve for $n$.

$(n-1) = \frac{90}{6} = 15 \Rightarrow n = 16$.

Here $a = 5$ and $d = 11 - 5 = 6$ (checked: $17 - 11 = 6$, $23 - 17 = 6$).

Set $t_n = 95$:

Since $n = 16$ is a positive integer, $95$ genuinely is a term of this AP, the $16$th.

Independent check: the $16$th term is $t_{16} = 5 + 15(6) = 5 + 90 = 95$ — confirms it.

Alternatively, count steps of $6$ from $5$ up to $95$: the gap is $95 - 5 = 90$, and $90 \div 6 = 15$ whole steps, so $95$ is $15$ steps after the first term — i.e. the $(15+1) = 16$th term.

Reporting $n = 15$ (the value of $n-1$) instead of $n = 16$ — always finish solving for $n$ itself. Also, if solving gave a fraction or non-integer, the correct conclusion would be that $95$ is not a term of the AP at all.