Q.Find the general (nth) term of the A.P. $5, 11, 17, 23, \ldots$, and determine whether $301$ is a term of this sequence. If it is, find its position.

[!TLDR] Use $a_n=a+(n-1)d$ with $a=5,d=6$; set it equal to $301$ and check whether $n$ comes out as a positive integer. [!ANSWER] $a_n=6n-1$; since $6n-1=301$ gives $n=50.33\ldots$, which is not an integer, $301$ is not a term of this A.P.

The A.P. is $5,11,17,23,\ldots$ with first term $a=5$ and common difference $d=11-5=6$. By the general-term formula, $a_n=a+(n-1)d=5+6(n-1)=6n-1$. To check whether $301$ is a term, set $a_n=301$: $6n-1=301 \Rightarrow 6n=302 \Rightarrow n=\dfrac{302}{6}=50.33\ldots$. Since a valid term-position $n$ must be a positive integer, and $50.33\ldots$ is not an integer, $301$ does not occur in this sequence. (As a check, $a_{50}=6(50)-1=299$ and $a_{51}=6(51)-1=305$, so $301$ falls strictly between two consecutive terms and is genuinely skipped.) [!ANSWER] $a_n=6n-1$; $301$ is not a term of the A.P.

Find $a$ and $d$, write the general term $a_n=a+(n-1)d$, then solve $a_n=$(given value) for $n$ — a term exists only if $n$ works out to a positive integer.

A common error is stopping at $n=50.33$ and rounding it to $n=50$ or $n=51$ and declaring the value a term anyway; a non-integer $n$ means the number is simply not in the sequence.