Middle Term

Let the smallest and largest numbers be $a$ and $b$ respectively. The common ratio can't be an integer (it's not allowed to be $1$ in the question, and anything else is too big to keep all the numbers with two digits), so it must be a fraction. Let the common ratio be $\frac{p}{q}$ , where $p$ and $q$ are integers with no common factor.

Then $b=\frac{p^4}{q^4} \cdot a$ ; so $q^4$ divides $a$ . Say $a=kq^4$ ; then the set of numbers is $kq^4,\;kpq^3,\;kp^2 q^2,\;kp^3 q,\;kp^4$

Since $b>a$ , we have $p>q>1$ . Note that $4^4=256$ is too big; so we can only have $p=3$ and $q=2$ . Since $3^4=81$ , we must have $k=1$ ; so the only valid sequence is $16,24,36,54,81$

or its reverse. Either way, the middle number is $\boxed{36}$ .

Let the first term and common ratio of the geometric progression be $a$ and $r$ respectively. Then the fifth term is $ar^4$ and $ar^4 < 100$ . Since the minimum two-digit $a$ is $10$ , then $r^4 < 10$ , $\implies 1 < r < \sqrt[4]{10} \approx 1.778$ .

Since the first five terms, $a, ar, ar^2, ar^3, ar^4$ , are integers, $r$ must be rational. Let $r = \frac pq$ , where $p$ and $q$ are positive integers. Then the fifth term $ar^4 = a \cdot \frac {p^4}{q^4}$ . For $ar^4$ to be an integer, $a$ must be an multiple of $q^4$ . Since $q$ cannot be $1$ , then $q \ge 2$ . If $q = 3$ , then the smallest $a=3^4 = 81$ . For a smallest $r = 1 \frac 13$ , then the second term $ar = 108$ , a three-digit number not acceptable. For $q > 3$ , $a > 100$ . Therefore, $q$ must be $2$ and $r = 1 \frac 12 = \frac 32$ . And the smallest $a=q^4 = 2^4 = 16$ and the first five terms are $\{16, 24, 36, 54, 81\}$ . The next $a = 32$ will have the fourth term $>100$ . Therefore there is only one geometric progression satisfying the conditions and the middle term is $\boxed{36}$ .