To solve a 7th-degree polynomial arising from a geometric progression

Question

I have a bit of difficulty with this maths problem, but not actually with solving it, but finding a proper (universal) way to solve it.

There is this geometric progression, where the first term is equal to 3, and the sum of first seven terms is equal to 381. What is the common ratio?

We can immediately see the ratio is not equal to one, so the formula for our sum will be $$S_n=a_1\cdot\frac{1-q^n}{1-q}$$ Now we can rearrange, substitute, etc. to get a polynomial like this: $$-q^7 + 127q - 126 = 0$$ $$-q(q^6 - 1) + 126(q - 1) = 0$$ $$(q - 1)(0 - q)(q + 1)(q^2 + q + 1)(q^2 - q + 1) + 126(q - 1) = 0$$ $$(q - 1)[-q(q + 1)(q ^ 2 + q + 1)(q^2 - q + 1) + 126] = 0$$ $q=1$ is a root, but it can't be the solution. And here I get stuck. I cannot see any way to extrapolate a common term (like I did with $q-1$).

WolframAlpha gives me the solutions as $1, 2$ and $\approx-2.38$, but its "show step-by-step solution" feature makes a rather big leap for the polynomial to roots. I found some site that explains how to solve such polynomials by trying to guess the roots and then dividing by them (e.g. try "1", if it is a root, divide by $q - 1$ and see what you get, and so on). This works here, because the root we are looking for is 2, but it's not a really universal solution.

Is there a proper way to solve this or do you really have to make guesses?

Answer 1

In general the equation becomes:

$$ S_n=a_1\frac{1-q^n}{1-q}=a_1 \frac{(1-q)(1+q+q^2+\cdots+q^{n-1})}{1-q} $$ that, for $a_1\ne 0$ and $q\ne 1$ becomes: $$ q^{n-1}+q^{n-2}+\cdots q+1-\frac{S_n}{a_1}=0 $$ This is an equation of degree $n-1$ and we know, from the Abel-Ruffini theorem, that a general solution in terms of radicals can be found only if the degree is $n-1<5$.

Answer 2

You are way overthinking this. Part of the problem is the question itself, it should have,specified that all the terms are positive. Otherwise the ratio is not unique.

Form your polynomial equation as above, and search for a rational root using the rational root theorem. You get two rational roots $q=1, q=2$. The former is wrong since it has a division by zero in the original formula. The latter works.

I am guessing that $2$ is the intended answer, not the other, irrational root that WA gives as about $-2.38$. The positive root must be unique since for positive common ratios, the sum is a monotonic function of the ratio.