The third term of a geometric sequence is 3 and the sixth term is 64/9. How do you find the fifth term of this sequence?

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Answer 1 · 2016-03-23T19:08:37

Write a systems of équations using the formula #t_n = a xx r^(n - 1)#

First equation:

#3 = a xx r^(3 - 1)#

Second equation:

#64/9 = a xx r^(6 - 1)#

Solve by substitution:

#3/r^2 = a -> 64/9 = 3/r^2 xx r^5#

#64/9 = (3r^5)/r^2#

We can simplify further using the exponent rule #a^x / a^m = a^(x - m)#

#64/9 = 3r^3#

#(64/9)/3 = r^3#

#64/27 = r^3#

#root(3)(64/27) = r#

#4/3 = r#

#3 = a xx (4/3)^2#

#3 = 16/9a#

#3 xx (9/16) = a#

#27/16 = a#

#t_n = a xx r^(n - 1)#

#t_5 = 27/16 xx (4/3)^4#

#t_5 = 27/16 xx 216/81#

#t_5 = 9/2#

The 5th term is #9/2#