Which term of the series $5,8,11,14,\dots$ is $320$?

The given series is an Arithmetic Progression (AP): $5,8,11,14,\dots$
Here, the first term $a=5$ and the common difference $d=8-5=3$.
Let the $n$ -th term be $320$. The formula for the $n$ -th term of an AP is:

$$T_n=a+(n-1)d$$

Substitute the values into the formula:

$$320=5+(n-1)3$$

$$320-5=3(n-1)$$

$$315=3(n-1)$$

$$n-1=\frac{315}{3}=105$$

$$n=105+1=106$$

Thus, $320$ is the 106th term of the series.