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}
Best Answer
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.