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