A sequence

Here the 1st term $a=5$ , diffrence $d=4$

So we know,

The sum is (for 1st 100th term): $S_{n}=\frac{n}{2} \left\{2a+(n-1)d \right\} \\\therefore S_{100}=\frac{100}{2} \left\{2\times5+(100-1)\times4\right\}\\ \quad\quad=50\times(10+99\times4)\\ \quad\quad=50\times{406}\\ \quad\quad=\boxed{\color{#20A900}20300}$

The $nth$ term of this sequence is $4n + 1$ , hence the $100th$ term is $401$ . As the difference between the numbers is always $4$ , we can pairs up the first and last numbers, e.g. $5 + 401 = 406$ . As there are $100$ terms, there are $\frac{100}{2} = 50$ lots of $406$ . $50 \times 406 = \boxed{20300}$ .