The sum of all those terms, of the arithmetic progression 3, 8, 13,...

Given sequence is $3, 8, 13, . . . . . 373$.

Here $a = 3, d = 5and{a}_{n} = 373$

Now ${a}_{n} = a + \left(n - 1\right)d$

$⇒ 373 = 3 + \left(n - 1\right)5$

$⇒ n = 75$

We know that ${S}_{n} = \dfrac{n}{2}\left(a + {a}_{n}\right)$

$⇒ {S}_{75} = \dfrac{75}{2}\left(3 + 373\right)$

$⇒ {S}_{75} = 14100$

Now let us write the sequence of terms which are divisible by $3$.

We get, $3, 18, 33, . . . . . 363$

$⇒ 363 = 3 + \left(n - 1\right)15$

$⇒ n = 25$

Now let us find the sum of terms divisible by $3$.

$⇒ {S}_{divby3} = \dfrac{25}{2}\left(3 + 363\right) = 4575$

Required sum =Sum of $75$ terms$-$Sum of terms divisible by $3$

$= 14100 - 4575$

$= 9525$.

Therefore, the required sum is $9525$