#a_n = a + (n1)d #
Here, #a# is the first term, #d# is the common difference and #a_n# is the nth term of the sequence.
We are given:

#a=2#

#a_4=16#

#a_n=11998#
#color(red)("To find "n" and " d.)#
Let's start with finding #d#:
#a_n = a + (n1)d #
#color(brown )("Put in "a=2 " and " n=4)#
#a_4 = 2 + (41)d #
#16 = 2 + (41)d # , # color(blue)(" since " a_4=16)#
#16 = 2 + 3d #
Add #2# to both sides:
#16 color(blue)(+ 2) = 2 color(blue)(+ 2)+ 3d #
#18 = 3d #
#color(red)(6 = d )#
Next, calculate #n#:
#a_n = a + (n1)d #
#color(brown)("Put in "a_n=11998 " , " a=2 " and " d=6)#
#11998 = 2 + (n1)6 #
Add #2# to both sides:
#11998 color(blue)(+ 2) = 2 color(blue)(+ 2) + (n1)6 #
#12000 = (n1)6 #
Divide both sides by #6#:
#12000/ color(blue)6 = [(n1)6]/ color(blue)6 #
#2000 = n1 #
Add #1# to both sides:
#2000 color(blue)(+ 1) = n1 color(blue)(+ 1) #
#color(red)(2001 = n) #