Rewriting $a,b,c,d,e$ as $a,ar,ar^2,ar^3,ar^4$, it can be seen that the $gcd(a,e)$ is $a$. Therefore, $a$ must be equal to $1$. However in the problem, it states that $a \ge 2$ so it is not possible for $a$ to equal $1$. Therefore, I believe that this question is flawed.

i think $a=k^{4}$ for some integer $k$ for there to be a solution because if you make $r$ a fraction with a denominator that is $k$ then you could eliminate all their common factors and have $\gcd(a,e)=1$. after very little inspection... you can see that $a=16$ is the only case that could possibly work and thus follows $r=\frac{3}{2}$ which would make $\boxed{c=36}$

Rewriting $a,b,c,d,e$ as $a,ar,ar^2,ar^3,ar^4$, it can be seen that the $gcd(a,e)$ is $a$. Therefore, $a$ must be equal to $1$. However in the problem, it states that $a \ge 2$ so it is not possible for $a$ to equal $1$. Therefore, I believe that this question is flawed.

Your assumption about $r$ being an integer is flawed. $\gcd(a,e)=a$ if $r$ is a nonzero integer