What is the 8th term of the geometric sequence where a1 = 625 and a3 = 25?

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Answer 1 · 2015-12-20T19:14:47

It could be #1/125# or #-1/125#

The general term of this sequence is of the form:

#a_n = a r^(n-1)#

where #a=625# is the initial term and #r# the common ratio.

Then:

#25 = a_3 = a r^2 = 625 r^2#

So #r^2 = 25/625 = 1/25#

That results in two possibilities for #r#, viz #r = 1/5# and #r = -1/5#

If #r = 1/5# then:

#a_8 = 625*(1/5)^7 = 5^(4-7) = 5^(-3) = 1/125#

If #r = -1/5# then:

#a_8 = 625*(-1/5)^7 = (-1)^7 5^(4-7) = -5^-3 = -1/125#