\(A\) is a geometric progression if and only if the ratios of adjacent terms, \(\frac{A_{i+1}}{A_i}\) \((1\leq i\leq N-1)\), are all equal;
this is equivalent to that

\[ \frac{A_{i+1}}{A_{i}}=\frac{A_{i+2}}{A_{i+1}} \]

holds for all \(1\leq i\leq N-2\).
However, this requires to compare fractions without errors.
Instead, we can even deform the equation as

\[ A_{i+1}^2=A_{i}A_{i+2} \]

to boil it down to comparisons of integers.
This can be implement with for and if statements.

Note that the terms \(A_i,A_{i+1},A_{i+2}\) are positive integers not exceeding \(10^9\), so the both hand sides do not exceed \(10^{18}\), which can be represented with \(64\)-bit integer types.
Note that using \(32\)-bit integer types like int in C++ causes overflows.

Sample code in C++:

#include <bits/stdc++.h>
using namespace std;

int main() {
	int n;
	long long a[100];
	bool flag=true;

	cin>>n;
	for(int i=0;i<n;i++){
		cin>>a[i];
	}
	for(int i=0;i<(n-2);i++){
		if((a[i]*a[i+2])!=(a[i+1]*a[i+1]))flag=false;
	}

	if(flag)cout<<"Yes"<<endl;
	else cout<<"No"<<endl;
}