Respuesta :
now, we know the 4th term is 5.... ok... now, what's the common difference? well, we dunno, but notice, from the 4th to the 13th term, you do have to use it 9 times to hop over to the 13th term, let's say is "d", then
[tex]\bf \begin{array}{llll} term&value\\ ------&------\\ a_4&5\\ a_5&5+d\\ a_6&(5+d)+d\\ a_7&(5+d+d)+d\\ a_8&(5+d+d+d)+d\\ a_9&(5+d+d+d+d)+d\\ a_{10}&(5+d+d+d+d+d)+d\\ a_{11}&(5+d+d+d+d+d+d)+d\\ a_{12}&(5+d+d+d+d+d+d+d)+d\\ a_{13}&(5+d+d+d+d+d+d+d+d)+d\\ &5+9d \end{array}[/tex]
we also know that the 13th term is -1
[tex]\bf \stackrel{a_{13}}{5+9d}=-1\implies 9d=-6\implies d=\cfrac{-6}{9}\implies \boxed{d=-\cfrac{2}{3}}[/tex]
now, recall above what the 8th term is
[tex]\bf a_8=5+4d\implies a_8=5+4\left(-\frac{2}{3} \right)\implies a_8=5-\cfrac{8}{3}\implies a_8=\cfrac{7}{3}[/tex]
[tex]\bf \begin{array}{llll} term&value\\ ------&------\\ a_4&5\\ a_5&5+d\\ a_6&(5+d)+d\\ a_7&(5+d+d)+d\\ a_8&(5+d+d+d)+d\\ a_9&(5+d+d+d+d)+d\\ a_{10}&(5+d+d+d+d+d)+d\\ a_{11}&(5+d+d+d+d+d+d)+d\\ a_{12}&(5+d+d+d+d+d+d+d)+d\\ a_{13}&(5+d+d+d+d+d+d+d+d)+d\\ &5+9d \end{array}[/tex]
we also know that the 13th term is -1
[tex]\bf \stackrel{a_{13}}{5+9d}=-1\implies 9d=-6\implies d=\cfrac{-6}{9}\implies \boxed{d=-\cfrac{2}{3}}[/tex]
now, recall above what the 8th term is
[tex]\bf a_8=5+4d\implies a_8=5+4\left(-\frac{2}{3} \right)\implies a_8=5-\cfrac{8}{3}\implies a_8=\cfrac{7}{3}[/tex]
