How do you write an nth term rule for #a_2=-30# and #a_5=3750#?

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Answer 1 · 2016-09-07T13:51:59

The general term is #t_n = 6 xx (-5)^(n - 1)#.

Write a system of equations in terms of #a# and #d#.

#-30 = a xx r^(2 - 1)#

#3750 = a xx r^(5 - 1)#

Solve by substitution.

#-30 = a xx r#

#a = -30/r#

#3750 = -30/r xx r^4#

#3750 = -30r^3#

#-125 = r^3#

#r = -5#

#:.-30 = a xx (-5)^1#

#-30 = -5a#

#a = 6#

#:.#The first term is #6# and the common ratio is #-5#.

The general term is given by #t_n = a xx r^(n - 1)#

#t_n = 6 xx (-5)^(n - 1)#

Hopefully this helps!