Respuesta :

Given:

[tex]t_n=\frac{1}{2}n^2+10[/tex][tex]\begin{gathered} 100=\frac{1}{2}n^2+10 \\ 90=\frac{1}{2}n^2 \\ 180=n^2 \\ n=\sqrt[]{180} \\ n=13.4164 \end{gathered}[/tex]

13 th term is closest to 100.