Numbers which cannot be formed We are given two numbers $a,b$ such that $a<b$. Now we have a set $\{a,a+1,a+2,\ldots, b\}$ (all number between a and b including them). Then, we have to find how many numbers cannot be formed from the above set. The only operation allowed on the set elements is addition.  
Note : We can add these number as many times we want. Only addition allowed on these numbers.
Eg : If numbers are 3 and 5, the set is $\{3,4,5\}$ using these we only cannot make $1,2$.
Can anyone help me with that?
I am not getting any idea. I tried the cases when $b-a = 1$. 

If you start with $a\neq b$ (i.e. if your starting set contains more than one number) then the set of numbers that "can be formed" will always be cofinal in the natural numbers (i.e. there will always be a $n$ s.t. every $m>n$ could be obtained via additions). 
To see this, assume you start from $\{n,n+1\}$. Notice that if you consider the possible results you can obtain by summing these two numbers $m$ times you obtain $m+3$ consecutive elements. Eg:
$$0:\quad \{n,n+1\} $$
$$1:\quad \{2n, 2n+1, 2n+2 \} $$
$$2:\quad \{3n, 3n+1, 3n+2, 3n+3 \} $$
and so on. This should come as no surprise as the number of $k$-combinations with repetition of $2$ elements is 
$$ \frac{(2+k-1)!}{(2-1)!k!} = \frac{(k+1)!}{k!}=k+1 $$
This implies that, after $m=n-1$ sums you have
$$ \{mn, mn+1, mn+2, ... , mn+n \} $$ 
and last term is exactly $(m+1)n$. 
Example:
start with $\{3,4\}$. Summing them once you get $\{6,7,8\}$. If you consider the possible results of summing $3$ elements drawn from $\{3,4\}$ you get $\{9, 10, 11, 12 \}$. But $12 = 3\cdot 4$, i.e. if you consider the  possible sums of $4$ elements drawn from  $\{3,4\}$ the smallest number you can get  is exactly $12$. This means that, from now on, you will be able to obtain every other number. In this example the only ones you cannot get are  $\{1,2,5\}$.
