A simple progression

This is the sum of an arithmetic progression with common difference, $d=3$ and first term, $a=1$

Now, we need to find the number of terms. Use the formula

$T_n = a + (n-1)(d)\\ 100 = 1 + (n-1)(3)\\ 3n = 102\\ n=34$

Then, use the formula to find the sum:

$S_n = \dfrac{n}{2} (a+T_n)\\ S_{34} = \dfrac{34}{2}(1+100) = 17(101) = \boxed{1717}$

$100=1+(n-1)*3$ ; n=34; S=((1+100)*34)/2=1717.

1 + 100 =101

4 + 97 = 101

...

49 + 52 = 101

There are 17 such pairs of terms. 17 * 101 = 1717

It follows an AP with first term(a) = $1$ and common difference(d) = $3$ . Let $100$ be $n$ the term $100 = 1 + (n - 1)×3\\ n = 34$ .
So, sum of all terms = $\dfrac{34}{2} × \left(1 + 100\right) = 17 × 101 = \color{#69047E}{\boxed{1717}}$ .