Let #n# be the position count within the sequence.
Let #a_n# be the #n^("th")# term
Given:
Term 1#->a_n->a_1=8 #
Term 2#->a_n->a_2=6#
Term 3#->a_n->a_3=4#
Term 4#->a_n->a_4=2#
Term 5#->a_n->a_5=0#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Observe the following relationship between terms
#a_1-a_2=8-6=2 #
#a_2-a_3=6-4=2#
#a_3-a_4=4-2=2#
#a_4-a_5=2-0=2#
So each consecutive term is 2 less than the previous term
Lets try and build a general equation relating it all to the first term of #a_1#
#color(green)("Term 1"->a_n->a_1=8 color(white)("d")color(red)(larr a_1-0))#
#color(green)("Term 2"->a_n->a_2=6color(white)("d")color(red)(larr a_1-2))#
#color(green)("Term 3"->a_n->a_3=4color(white)("d")color(red)(larr a_1-2-2")#
#color(green)("Term 4"->a_n->a_4=2color(white)("d")color(red)(larr a_1-2-2-2))#
#color(green)("Term 5"->a_n->a_5=0 color(white)("d") color(red)(larr a_1-2-2-2-2))#
#color(purple)("By counting the 2's we have for any term "a_n)#
#color(purple)(a_n=a_1-2xx(n-1) color(white)("ddd")color(red)( -> color(white)("ddd")a_1-2(n-1)))#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So we plot
#color(green)(color(white)("dddddddddddddddddd")"Actual relationship"color(white)("ddd")color(red)("xy graph equivalent"))#
#color(green)("Term 1"->a_n->a_1=8 -> obrace((n,a_1)=(1,8))color(white)("d")color(red)(obrace(larr(x,y)=(1,8))))#
#color(green)("Term 2"->a_n->a_2=6-> (n,a_2)=(2,6)color(white)("d")color(red)(larr(x,y)=(2,6)))#
#color(green)("Term 3"->a_n->a_3=4->(n,a_3)=(3,4)color(white)("d")color(red)(larr(x,y)=(3,4)))#
#color(green)("Term 4"->a_n->a_4=2->(n,a_4)=(4,2)color(white)("d")color(red)(larr(x,y)=(4,2)))#
#color(green)("Term 5"->a_n->a_5=0->(n,a_5)=(5,0)color(white)("d")color(red)(larr(x,y)=(5,0)))#
