Common ratio of a geometric progression

Question

In a geometric progression containing $6$ terms, the first term is $2$ and the sum of $6$ terms is $728$.

What is the common ratio?

Answer 1

Hint. By definition of geometric progression, the sum of the six terms is given by $$2+2k+2k^2+2k^3+2k^4+2k^5=728\implies k(1+k+k^2+k^3+k^4)=\frac{728-2}{2}=363$$ where $k$ is the common ratio. Can you take it from here?

P.S. Note that the equation $k(1+k+k^2+k^3+k^4)=363$ has a unique positive solution (the polynomial on the left is strictly increasing for $k> 0$). Try with some integer value.

Answer 2

The partial sum $$2\sum_{k=0}^{5}x^{k-1}$$ can be simplified to

$$2\cdot \frac{1-x^6}{1-x}=728$$

$$ \frac{1-x^6}{1-x}=364$$

This is a polynomial equation

$x^6-364x+363=0$

Now apply the Rational root theorem. For this purpose you have to factorize $363$. After you have found a solution you have to check it.