Question

If $(1-p)(1+3x+9x^2+27x^3+81x^4+243x^5)=1-p^6,p\ne 1$, then the value of $\frac{p}{x}$ is

Answer 1

The given equation can be rewritten as follows:
$1+3x+9x^2+27x^3+81x^4+243x^5=\frac{1-p^6}{1-p}$
The LHS can be simplified as:
$1+3x+9x^2+27x^3+81x^4+243x^5=1+{\left(3x\right)}^1+{\left(3x\right)}^2+{\left(3x\right)}^3+{\left(3x\right)}^4+{\left(3x\right)}^5$
The LHS is the sum of 6 terms which are in Geometric Progression (G.P.) with the common ratio $3x$ and the first term being 1.
The sum of $n$ terms in G.P. with the common ratio $r$ and first term $a$ is $\frac{a(1-r^n)}{1-r}$
Thus, the LHS reduces to $\frac{1-{\left(3x\right)}^6}{1-3x}$
On comparing LHS and RHS, we observe that $3x=p\Rightarrow \frac{p}{x}=3$