Geometrical Sequence: Four

Relevant wiki: Arithmetic Progressions

$4+8+12+...+100=4(1+2+3+...+25)$

By Arithmetic Progressions $1+2+3+...+25=\frac{(25)(26)}{2}=325$

$\therefore 4(1+2+3+...+25)=4(325)=1300$

1288 is resoult

$4 + 8 + 12 + \ldots + 100$ is a sum of an arithmetic progression where

$a=4 \quad d=8-4=4 \quad T_n = 100$

We can use the formulas for $T_n$ and $S_n$ for arithmetic progressions to solve this.

$T_n = a + (n-1)(d) \implies 100 = 4 + (n-1)(4) \implies n=25$

Therefore,

$4+8+12+\ldots+100 = S_{25} = \dfrac{25}{2}\left(4 + 100\right) = \boxed{1300}$