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A geometric progression has 625 as the first term. The product of its first 3 terms is equal to the product of its first 6 terms. Find the common ratio of the progression.

Rodrigo Vu
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    @YvesDaoust I think someone thinks this question, in its current form, doesn't deserve answers. And they might be right, as per our community guidelines. As for downvoting answers because of that, I personally wouldn't go that far. But I have seen it happen before. – Arthur Jun 05 '20 at 08:18
  • @Arthur: vote for closing and downvote of the question are logical, but always better to explain. –  Jun 05 '20 at 08:23
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    @YvesDaoust I agree. In this case, I haven't touched any voting, but I always explain myself when I do (or upvote a comment that already explains). – Arthur Jun 05 '20 at 08:24
  • @Arthur: on second thoughts, the downvote is well deserved, but not the closing. The question is clear enough and interesting. (I am stopping here.) –  Jun 05 '20 at 08:26
  • @YvesDaoust Yves Daoust: $1877$ up-votes, $3562$ down-votes, talking about downvoting... ;-) – Déjà vu Jun 05 '20 at 10:39
  • @e2-e4: in what way would my personal "score" be relevant to this case ? –  Jun 05 '20 at 10:48

1 Answers1

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Write the terms of the progression as $a_n = 625 r^n$. This gives that our first term, $a_0$, is $625 = 5^4$; call this $a$. The product of the first three terms is hence $a \cdot ar \cdot ar^2 = a^3r^3$, and similarly, we have the product of the first six terms is $a^6r^{15}$. Thus either $r=0$ or $$a^3 r^3 = a^6 r^{15},$$ giving $a^3r^{12} = 1$, or $5^{12} r^{12} = 1$.

So $r = \pm\frac 15$ or $0$.

edit as Yves correctly points out, $r=0$ isn't actually allowed in the definition of a GP, so unless otherwise specified, the answer is $\pm\frac 15$.

hdighfan
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  • My bad; I've fixed it now. (This problem doesn't seem like the type of thing to require complex numbers, so I've ignored those solutions for now and worked over reals.) – hdighfan Jun 05 '20 at 08:19
  • In principle $r=0$ is not allowed for a geometric progression. –  Jun 05 '20 at 08:20
  • You can save yourself a little bit of work because the condition for the product of the first six terms versus the product of the first three terms tells us immediately that $ (ar^3) \ · \ (ar^4) \ · \ (ar^5) \ = \ 1 \ \ , $ bringing you immediately to your result $ \ r^4 \ = \ \frac{1}{a} \ = \ \frac{1}{625} \ \ . $ –  Jul 30 '21 at 06:51