Ellipse representation 
The equation $\frac{x^2}{2-a}+\frac{y^2}{a-5} +1 = 0$ represents an ellipse if $a\; \epsilon$
  (A) $(2,\frac{3}{2})\;\cup\;(\frac{3}{2},5)$
  (B) $(2,\frac{3}{2})$
  (C) $(1,\frac{3}{2})$
  (D) $(\frac{3}{2},5)$

This is what I have done,
For representing an ellipse,
 $$e<1$$
 $$\implies \sqrt{1-\frac{a-5}{2-a}} <1$$
 $$\implies 1-\frac{a-5}{2-a}<1$$
 $$\implies \frac{a-5}{2-a}>0$$
 $$\implies \frac{a-5}{a-2}<0$$  
So, $a\;\epsilon\;(2,5)$
But there are no such options. So, have I missed a solution ? I think there is a mistake in the way I have solved the problem. Can anybody tell me what to do?
 A: you should exclude the case of $e=0$, which is a circle but not a ellipse. So the answer is A.
Additional notes
I think the $3/2$ in all your options should be $7/2$. Otherwise $(2,3/2)$ does not make any sense. 
A: If
$$\frac {x^2}{2-a}+\frac{y^2}{a-5}+1=0$$
is claimed to be an ellipse, then in standard form,
$$\frac {x^2}{a-2}+\frac{y^2}{5-a}=1$$
which matches exactly with the result that you found, namely that $a\in(2,5)$.
The listed answer in $(A)$ is not correct as written, but the writing of it suggests that perhaps it is written incorrectly and might be something like $(2,\frac 52)\cup (\frac 52,5)$.  But if that were the case, then (presumably) $(B),(D)$ would also have the same mixup and would also be correct answers.
Basically, you found the correct interval for $a$, I would go back to the source of the question and make sure that nothing is misprinted or mis-copied...
A: If we agree in $(2,3/2)=\emptyset$ and notice that there is an if and no iff, then there are two interpretations depending on we consider a circle as an ellipse or not. In the first case A, C, and D are correct, otherwise C is the only correct solution.  
