Proving that this set A can be written as $A= x_0 + W$ for some $x_0 \in V $ and some sunspace W of V, 
Question: Let $V$ be a finite dimensional vector space over $\mathbb{R}$. Suppose that a subspace $A\subseteq V$ has the following property: For any finite set of scalars $a_1 , a_2 ,\dots, a_n \in  \mathbb{R}$ which satisfies $\sum_{i=1}^n a_i = 1$ and any vectors $v_1, \dots, v_n \in A$, $ a_1 v_1 + \dots + a_n v_n  \in A $. Then show that $A=x_0 +W$ for some $x_0 \in V$ and some subspace $W$ of $V$, where $x_0 + W = \{ x_0 + v : v \in  W\}$.

It was asked in my Linear Algebra quiz ( Over two hours ago).
I understood what is to be done but there is unfortunately no intuition on how should I approach this and so I would very much appreciate any hints.
I have been following Linear Algebra by  Hoffman and Kunze if that helps.
 A: Yes, the problem is misstated, as @JoséCarlosSantos pointed out. $A$ is called an affine subspace, but not in general a subspace. Of course, we also need to be sure that $A$ is non-empty. As I suggested in my comment, you have to apply the given hypothesis with both $n=2$ and $n=3$.
Choose any $x_0\in A$, and consider $W=A-x_0 = \{w\in V: w=v-x_0 \text{ for any }v\in A\}$.
We claim first that $W$ is closed under addition. Choose $w_1,w_2\in W$ arbitrary. Set $v_1=w_1+x_0$, $v_2=w_2+x_0$; these are in $A$. Then (using the hypothesis with $n=3$) we have
$$-x_0+v_1+v_2 \in A$$
(since $-1+1+1=1$), and therefore
$$w_1+w_2 = (v_1-x_0) + (v_2-x_0) \in A-x_0 = W.$$
Now, let $c$ be an arbitrary scalar. Given $w\in W$, we wish to show that $cw\in W$. Set $v=w+x_0\in A$. Then (now using the hypothesis with $n=2$),
$$cw+x_0 = c(v-x_0)+x_0 = cv + (1-c)x_0 \in A,$$
(since $c+(1-c)=1$). We conclude that $cw\in W = A-x_0$.
Thus, $W$ is indeed a subspace.
(As one who taught linear algebra for over 35 years, I would comment that this is a rather a challenging quiz problem unless students have confronted the main ideas in homework prior to the quiz. The fact that it is not made explicit that $n$ can be arbitrary is troubling to me. In particular, the result cannot be deduced, I believe, just knowing the hypothesis for $n=2$. So if one assumes $n$ is fixed and arbitrary, one must know that $n\ge 3$.)
A: If the set A has only one element then W = {0}. If it has more than one then  fix $x_0 \in A$ and $y_0 \in A$ and define $w_0 = x_0-y_0$. One has to show that $A -w_0$ is a vector space. Take $x_1,...x_n \in A-w_0$ and numbers $b_1,...,b_n$ such that $\sum b_k$ is not $0$. Define $a_k = \frac{b_k}{\sum b_n}$. Clearly $\sum a_k = 1$. Now, each $x_k$ has form $y_k-w_0$ for some $y_k \in A$. By assumption: $$\sum_k a_ky_k$$ is in A, say the sum is $y$. Then $$\sum b_kx_k = (\sum b_n)\sum \frac{b_k}{\sum b_k} x_k = (\sum b_n)\sum \frac{b_k}{\sum b_k} (y_k-w_0) = (\sum b_n)(y-w_0)$$
So, to finish, we have to show that if z is in $A-w_0$ then $bz$ is in $A-w_0$. But this is the same as to show that if $z = x- w_0$ then $bz$ has format $y - w_0$ for some $y \in A$. Define this $y$ to be $bx-(b-1)w_0$. Then $y \in A$ since $b -( b-1) = 1$.
What is left is to consider case of $\sum b_k = 0$. But since the spaces are finite dimensional this follows from compactness argument.
A: This is not an answer, just too long for a comment.
(Certainly not something I would expect in a quiz.)
It is sufficient to take $n=2$ above. Suppose this is given.
Let $N \ge 1$ and let $P_N$ be the proposition that for any scalars $a_1,...,a_N$ that sum to 0ne and for any $v_1,...,v_N \in A$ then $\sum_{k=1}^N a_k v_k \in A$. $P_1 $ is trivial, and we are assuming $P_2$.
Suppose $N \ge 2$.
Suppose $a_1,...,a_N,a_{N+1}$ sum to one and $v_1,...,v_N, v_{N+1} \in A$.
Let $v= \sum_{k=1}^{N+1} a_k v_k$.
If any of the $a_1,...,a_N$ are zero then we reduce to $P_N$, so we may assume that all $a_k \neq 0$.
Suppose $a_{N+1} \neq 1$, then
$v = (1-a_{N+1}){\sum_{k=1}^N a_k v_k \over \sum_{k=1}^N a_k } + a_{N+1} v_{N+1}$, then since ${\sum_{k=1}^N a_k v_k \over \sum_{k=1}^N a_k } \in A$ by $P_N$, we see that $v \in A$ by $P_2$.
Now suppose $a_{N+1} = 1$, then since $\sum_{k=1}^N a_k = 0$, and we have assumed the $a_k \neq 0$, at least one of the $a_1,...,a_N$ must be strictly positive. Without loss of generality suppose $a_1>0$ and so $a_1+1 > 1$. Write
$v= (-a_{1}){\sum_{k=2}^N a_k v_k \over \sum_{k=2}^N a_k } + (a_1+1) { (a_1v_1+v_{N+1}) \over a_1+1}$ and as above we can reduce to $P_{N-1},P_2$
