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:

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