For any term in an AP, the rule is the same: #T_n = a+(n-1)d#
For the #6th# term we have..
#T_6 = a+ 5d =-31" "larr (6-1=5)#
For the #14th# term we have:
#T_14 = a+13d =-135" "larr(14-1 =13)#
To be able to write the rule we need to find #a# and #d#.
Solve the equations simultaneously:
#color(white)(xxxxxx)T_6 = a+" "5d=-31#.........................................A
#color(white)(xxxxxx)T_14 = a+ " "13d=-135#.........................................B
A-B:#color(white)(wwwwwwwwww)-8d = 104#
A-B:#color(white)(wwwwww.wwwwww)d = -13#
Subst in A:#rarra +5(-13) = -31#
#" "a" " -65 = -31#
#" "a " "= -31+65#
#" "a " "= 34#
Now we know that the first term is #34# and that #d=-13#
The first #6# terms are: #34, " "21," " 8, -5, -18, -31 .....#
The rule for the #nth# term is:
#T_n = 34+(n-1)(-13)#
#T_n = 34-13n +13#
#T_n = -13n +47#
