Each consecutive term in the sum is separated by a difference of 6, so the [tex]n[/tex]-th term is [tex]4+6(n-1)=6n-2[/tex] for [tex]n\ge1[/tex]. The last term is 70, so there are [tex]6n-2=70\implies n=12[/tex] terms in the sum.
Now,
[tex]S=4+10+\cdots+64+70[/tex]
but also
[tex]S=70+64+\cdots+10+4[/tex]
Doubling the sum and grouping terms in the same position gives
[tex]2S=(4+70)+(10+64)+\cdots+(64+10)+(70+4)=12\cdot74[/tex]
[tex]\implies\boxed{S=444}[/tex]
