What is the 15th term of the geometric sequence 52, 39, 29.25 …?

1 Answer
George C.
Oct 24, 2015

#(3/4)^14 * 52 ~~ 0.9265#

Explanation:

The first term is #a_1 = 52# and the common ratio is #3/4#:

#39/52 = (3*13)/(4*13) = 3/4#

#29.25/39 = (9.75*3)/(9.75*4) = 3/4#

So in general we can write:

#a_n = r^(n-1) a_0 = (3/4)^n*52#

So we find:

#a_15 = r^14 a_0 = (3/4)^14 * 52 = (3^14 * 52)/4^14#

#=(4782969 * 52) / 268435456 = 248714388 / 268435456#

#~~ 0.9265#

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