Respuesta :
as you may already know, we get an arithmetic sequence by adding a number or common difference to the current to get the next one, or "d".
we know the 3rd term is 11, or a₃ = 11, we also know that the 6th term is -1, or a₆ = -1, but we also know that to get from the a₃ to a₆ we had to use "d" several times, let's see
[tex]\begin{array}{rll} term&value\\ \cline{1-2} a_3&11\\ a_4&11+d\\ a_5&(11+d)+d\\ a_6&(11+d+d)+d\\ &11+3d\\ &-1 \end{array} \implies 11+3d=-1\implies 3d=-12\implies d = -4[/tex]
well, now we know that d = -4, hmmmm what's the first term anyhow, well, let's say that the first term has a value of F, or a₁ = F, so let's use another table for it like before from a₁ to a₃
[tex]\begin{array}{rll} term&value\\ \cline{1-2} a_1&F\\ a_2&F+d\\ a_3&(F+d)+d\\ &F+2d\\ &11 \end{array} \implies \begin{array}{llll} F+2d = 11\implies F+2(-4)=11\\\\ F-8=11\implies F = 19 \end{array}[/tex]
well, since now we know what's a₁ and "d", let's use them both
[tex]n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} a_n=n^{th}\ term\\ n=\stackrel{\textit{term position}}{27}\\ a_1=\stackrel{\textit{first term}}{19}\\ d=\stackrel{\textit{common difference}}{-4} \end{cases} \\\\\\ a_{27}=19+(27-1)(-4)\implies a_{27}=19+(-104)\implies \boxed{a_{27}=-85}[/tex]
