Question #47674

1 Answer
Oct 27, 2017

#108#

Explanation:

for any number in a geometric progression:
#u_n = a * r^(n-1)#

#r#: common ratio [a number that one number in a sequence is multiplied by to get to the next]

#a#: starting number #(u_1)#

#u_n#: a number at a certain point #n# in the sequence

#u_1 = 4#
#u_3 = 36#

#36/4 = 9#
#u_3/u_1 = 9#

#a_3/a_1 = r^2# (two positions apart, so the number is multiplied by #r# twice)

#r^2 = 9#

#r = 3#

then plug this value #r# into the equation:
#u_n = a * 3 *n^-1#

for the #4#th term, #n = 4#

#u_4 = a * 3^3#
#u_4 = 27a#

#a# is the first term of the progression, which, here, is #4#

#u_4 = 27 * 4#

#u_4 = 108#

the #4#th term of the sequence is #108#.