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

Given sequence is $3,8,13,.....373$.

Here $a\mathit{=}3\mathit{,}\mathit{ }d\mathit{=}5\mathit{ }\mathrm{and}\mathit{ }{a}_n\mathit{=}373$

Now $a_n=a+\left(n-1\right)d$

$\Rightarrow 373=3+\left(n-1\right)5$

$\Rightarrow n=75$

We know that $S_n=\frac{n}{2}\left(a+a_n\right)$

$\Rightarrow S_{75}=\frac{75}{2}\left(3+373\right)$

$\Rightarrow S_{75}=14100$

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

We get, $3,18,33,.....363$

$\Rightarrow 363=3+\left(n-1\right)15$

$\Rightarrow n=25$

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

$\Rightarrow S_{div by 3}=\frac{25}{2}\left(3+363\right)=4575$

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

$=14100-4575$

$=9525$.

Therefore, the required sum is $9525$