I'm able to get to an inequality for the sum of the arithmetic sequence greater than the sum of the geometric sequence, and have solved the inequality by guess and check and have verified the inequality by inputting the answer, however, I am struggling to find a formal algebraic proof to the problem...Any help is appreciated!
Hint: The sum of the first $n$ terms of an arithmetic series is $$a_0 +a_1+\cdots+a_{n-1} =a_0+(a_0+d)+\cdots+(a_0+(n-1)d)$$ $$=na_0+(1+2+\cdots+(n-1))d =\boxed{na_0+\left(\dfrac{ (n-1)n}{2}\right)d}$$
The sum of the first $n$ terms of a geometric series (assuming $r\neq 1$ -- otherwise it's just $ng_0$) is $$g_0+g_1+\cdots g_n=g_0 + g_0r+\cdots g_0r^{n-1}$$ $$=g_0(1+r+\cdots+r^{n-1}=\boxed{g_0\left(\dfrac{1-r^n}{1-r}\right)}$$
Plug in your specific values for the constants ($a_0,d,g_0,r$), set the two equal to each other, and solve for $n$ (it will be a quadratic equation in $n$). The next higher integer will be your answer (the solution the the quadratic may or may not be an integer, but you want the next integer in either case). You may have to decide if one of your solutions to the quadratic is meaningless (for example, you need $n\geq1$).