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There is a question in my book stating that

A geometric progression consists of an even number of terms. If the sum of all the terms is five times the sum of terms occupying odd places then find the common ratio.

I solved it correctly but I want to ask what if we have an odd number of terms and rest of the data are left undisturbed?

I tried but couldn't find the solution because in the earlier problem the number of odd terms is exactly half of the total terms but in this new part it is one more than the number of terms on even place.

My attempt :

There are for sure one more number at odd place than at even place . So first term is $a$, the common ratio is $r^2$ and the number of terms is $n+1$ ( if total number of terms in the original G.M. are $2n+1$).

So I did this but couldn't proceed further

\begin{align} \frac{r^{2n+1}-1}{r-1} &= 5\frac{(r^2)^{n+1}-1}{r^2-1}\\ (r+1)(r^{2n+1}-1)&= 5(r^{2n+2}-1)\\ r^{2n+2} -r + r^{2n+1}-1 &=5r^{2n+2}-5 \\ 4r^{2n+2} -r^{2n+1}+r-4 &= 0. \end{align}

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    $\begingroup$ First term $a$, common ratio $r$, $2n+1$ terms, sum is $a{r^{2n+1}-1\over r-1}$. The terms at the odd places are (in one interpretation) $$a+ar^2+ar^4+\cdots+ar^{2n}=a{r^{2n+2}-1\over r^2-1}$$ So set one equal to six times the other and see what happens. $\endgroup$ Aug 9 '20 at 7:16
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    $\begingroup$ Any thoughts on my comment, Ankit? $\endgroup$ Aug 10 '20 at 12:46
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    $\begingroup$ Just to say: despite what you claim in the last sentence, there are always one more term in an odd position then there are terms in even positions. If there are $2k+1$ terms in total, there are $k+1$ in odd positions and $k$ in even positions. $\endgroup$
    – lulu
    Aug 11 '20 at 18:36
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    $\begingroup$ I think this question is in danger of being closed for a second time. Please edit your post to include some effort. The comment from @GerryMyerson is pretty close to a full solution...why not give it a try? $\endgroup$
    – lulu
    Aug 11 '20 at 20:43
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    $\begingroup$ @lulu: The user was online 6 hours ago, he is from India, and in India it is now 2:30. Maybe we give him a short break before we knock him down. $\endgroup$
    – miracle173
    Aug 11 '20 at 21:12
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This is a case where a seemingly small change in the statement of a problem has a big effect on the ease with which the problem can be solved.

The original problem was to solve the equation

$$(1+r+r^2+\cdots+r^{2n-1})=5(1+r^2+r^4+\cdots+r^{2n-2})$$

The OP's variant is to solve the equation

$$(1+r+r^2+\cdots+r^{2n})=5(1+r^2+r^4+\cdots+r^{2n})$$

In the original problem, a minor miracle takes place, making the equation easy to solve: the variable $n$ drops out when you write $1+r+r^2+\cdots+r^{2n-1}$ as $r^{2n}-1\over r-1$ and $1+r^2+r^4+\cdots+r^{2n-2}$ as $(r^2)^n-1\over r^2-1$. But this doesn't occur in the variant. Instead we are left with the unwieldy polynomial

$$4r^{2n}-r^{2n-1}+4r^{2n-2}-r^{2n-3}+\cdots+4r^2-r+4=0$$

This polynomial clearly has no roots with $r\lt0$. By writing it as

$$4+r(4r-1)(1+r^2+r^4+\cdots+r^{2n-2})=0$$

we can easily see that there are no roots with $r\gt1/4$ either. Finally, for $0\le r\le1/4$ we have

$$4-r+4r^2-r^3+\cdots-r^{2n-1}+4r^{2n}\ge4-4r+4r^2-4r^3+\cdots-4r^{2n-1}+4r^{2n}\ge4(1-r)\ge3\gt0$$

so the polynomial of degree $2n$ has no real roots at all. It does, of course, have $2n$ complex roots (in $n$ pairs of complex conjugates). If $n=1$, for example, we have $4r^2-r+4=0$, with roots

$$r={1\pm3\sqrt{-7}\over8}$$

For $n\gt1$ the complex roots may or may not have nice radical expression; my guess is not.

Remark: The OP is to be commended for going beyond the original, simple question and asking about a variant. It's the hallmark of an inquiring, mathematical mind.

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  • $\begingroup$ Gerry Myerson points out in comments beneath the OP that "five times more than the sum" means that one expression is six times the other, not five. If indeed that's what the OP means, then all my $5$'s should be $6$'s, and all my $4$'s should be $5$'s. The basic answer still stands, though. I'll change things if/when the OP confirms Gerry's interpretation. $\endgroup$ Aug 11 '20 at 22:44
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    $\begingroup$ +1 This is much better than what I intended to write up. $\endgroup$
    – hardmath
    Aug 11 '20 at 23:31
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    $\begingroup$ @Ankit, by subtracting the left hand side of $(1+r+r^2+\cdots+r^{2n})=5(1+r^2+r^4+\cdots+r^{2n})$ from the right hand side (and writing the powers in reverse order, from largest to smallest). $\endgroup$ Aug 12 '20 at 10:20
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    $\begingroup$ @Ankit, I would also recommend you begin teaching yourself the rudiments of TeX, to make your math expressions easier to read. There's a bit of a learning curve to climb in doing so, but you'll find the view from the mountaintop worth the ascent. TeX is part of nearly every mathematician's toolbox. $\endgroup$ Aug 12 '20 at 11:37
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    $\begingroup$ Here is help with formatting math on this site, Ankit: math.meta.stackexchange.com/questions/5020/… $\endgroup$ Aug 12 '20 at 12:15

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