Question

Escreva a PA de quatro termos, em que o 1º- termo é a1 = 26 e a razão é r = 8.

Answer 1

$a_1=26\\ r=8$

$\boxed{a_n=a_k+(n-k)r}$

$a_2=a_1+r=26+8=34\ \therefore\ \boxed{a_2=34}\\\\ a_3=a_2+r=34+8=42\ \therefore\ \boxed{a_3=42}\\\\ a_4=a_3+r=42+8=50\ \therefore\ \boxed{a_4=50}$

$\boxed{(a_1,a_2,a_3,a_4)=(26,34,42,50)}$