Question

Determine uma PA de quatro termos, sabendo que a soma dos dois primeiros é 29 e dos dois últimos é 89. (8,21,36,51) (3,28,43,46) (7,22,36,51) (7,22,37,52) (9,20,35,54)

Answer 1

$\mathrm{PA=(a_1,a_2,a_3,a_4)=(a_1,a_1+r,a_1+2r,a_1+3r)}\\\\ \mathrm{a_1+a_2=29\ \to\ a_1+a_1+r=29\ \to\ \boxed{\mathrm{2a_1+r=29}}\ \mathbf{(I)}}\\\\ \mathrm{a_3+a_4=89\ \to\ a_1+2r+a_1+3r=89\ \to\ \boxed{\mathrm{2a_1+5r=89}}\ \mathbf{(II)}}\\\\ \mathbf{*\ Somando\ II\ com\ (-1).I,\ vem:}\\\\ \mathrm{2a_1+5r-2a_1-r=89-29\ \to\ 4r=60\ \to\ \boxed{\mathrm{r=15}}}\\\\ \mathbf{*\ Substituindo\ o\ valor\ de\ r\ em\ I,\ vem:}\\\\ \mathrm{2a_1+15=29\ \to\ 2a_1=29-15\ \to\ 2a_1=14\ \to\ \boxed{\mathrm{a_1=7}}}$

$\mathrm{PA=(7,7+15,7+2.15,7+3.15)}\\\\ \boxed{\boxed{\mathbf{PA=(7,22,37,52)}}}$