Let 3, 7, 11, 15, . . , 403 and 2, 5, 8, 11, . . . , 404 be two...

Given AP are: $3, 7, 11, 15, . . , 403\text{and}2, 5, 8, 11, . . . , 404$

${a}_{1} = 3, {a}_{2} = 2, {d}_{1} = 4, {d}_{2} = 3$

So, for the common AP,

$a = 11, d = LCM\left(3, 4\right) = 12$

Also, ${a}_{n} \le 403$

$⇒ 11 + \left(n - 1\right)12 \le 403$

$⇒ - 1 + 12n \le 403$

$⇒ n \le \dfrac{404}{12}$

$⇒ n \le 33 . 6$

$⇒ n = 33$

$⇒ {S}_{n} = \dfrac{33}{2}\left[2 \times 11 + 32 \times 12\right]$

$⇒ {S}_{n} = 6699$