Question

calcule a soma dos 28 termos iniciais da p.a. (2,14,26,...326

Answer 1

$\mathrm{PA=(2,14,26,\dots,326)}\\\\ \mathrm{r=a_2-a_1\ \to\ r=14-2\ \to\ r=12}\\\\ \mathrm{a_n=a_k+(n-k)r\ \to\ a_{28}=a_1+(28-1).r}\\ \mathrm{a_{28}=2+27.12\ \to\ a_{28}=2+324\ \to\ a_{28}=326}\\\\ \mathrm{S_n=\cfrac{(a_1+a_n).n}{2}\ \to\ S_{28}=\cfrac{(a_1+a_{28}).28}{2}}\\\\ \mathrm{S_{28}=\cfrac{(2+326).28}{2}=328.14}\ \to\ \mathbf{S_{28}=4592}$