Respuesta :

[tex]n=8\\ a_1=2\cdot3+3=9\\ a_n=a_8=2\cdot10+3=23\\\\ \displaystyle \sum\limits_{i=3}^{10}2i+3=S_8=\dfrac{8}{2}\cdot(9+23)=4\cdot32=128[/tex]