prove that, for some $p$ & $q, a_p, a_p+a_{p+1} + \cdots$ all are positive Let $ a_1,a_2,\ldots ,a_{100}$ be real numbers, each less than one, satisfy
 $ a_1+a_2+\cdots+a_{100} > 1$
Show that there exist two integers $p$ and $q$ , $p<q$, such that the numbers
$$a_q, a_q+a_{q-1}, \ldots, a_q+\cdots+a_p,$$
$$a_p, a_p+a_{p+1},\ldots,a_p+\cdots+a_q$$
are all positive.
I proved that if $n$ is the smallest integer such that 
$ a_1+a_2+\cdots+a_n>1$ 
then all the sums $a_n,a_n+a_{n-1},\ldots,a_n+\cdots+a_1$ are positive.
Will this help to prove?
source: Test of Math at 10+2 level( A collection of old ISI B.stat & B.math entrance exam question papers)
 A: [I found this solution collaboratively with someone else offline.]
$\def\nn{\mathbb{N}}$
$\def\rr{\mathbb{R}}$
Let $T(n) = ( \text{The theorem is true for any length-$n$ sequence from $\rr$} )$, for any $n \in \nn$.
If $T(n)$ is false for some $n \in \nn$:
  Let $m \in \nn$ be the minimum such that $T(m)$ is false [by well-ordering].
  Let $a_{1..m}$ be a sequence that does not satisfy the theorem.
  For any $p \in [1..m]$:
    If $\sum_{k=1}^p a_k \le 0$:
      $\sum_{k=p+1}^m a_k \ge \sum_{k=1}^m a_k > 1$.
      Also $a_{p+1..m}$ satisfies the theorem [by minimality of $m$].
      Thus some segment of $a_{p+1..m}$ has initial and terminal segments all with positive sum.
      But any segment of $a_{p+1..m}$ is also a segment of $a_{1..m}$.
      Thus $a_{1..m}$ satisfies the theorem, which gives a contradiction.
    Therefore $\sum_{k=1}^p a_k > 0$.
    Similarly $\sum_{k=p}^m a_k > 0$.
  Therefore $a_{1..m}$ has initial and terminal segments all with positive sum.
  Thus $a_{1..m}$ satisfies the theorem, which gives a contradiction.
Therefore $T(n)$ is true for any $n \in \nn$.
A: If the sum  is positive, at least one element is positive. Pick that one as "sequence" (of length 1).
