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} &gt; 1$

Show that there exist two integers $p$ and $q$ , $p&lt;q$, 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&gt;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 &amp; B.math entrance exam question papers)
[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 $n \in \nn$:

&emsp; Let $m \in \nn$ be the minimum such that $T(m)$ is false [by well-ordering].

&emsp; Let $a_{1..m}$ be a sequence that does not satisfy the theorem.

&emsp; For any $p \in [1..m]$:

&emsp; &emsp; If $\sum_{k=1}^p a_k \le 0$:

&emsp; &emsp; &emsp; $\sum_{k=p+1}^m a_k \ge \sum_{k=1}^m a_k &gt; 1$.

&emsp; &emsp; &emsp; Also $a_{p+1..m}$ satisfies the theorem [by minimality of $m$].

&emsp; &emsp; &emsp; Thus some segment of $a_{p+1..m}$ has initial and terminal segments all with positive sum.

&emsp; &emsp; &emsp; But any segment of $a_{p+1..m}$ is also a segment of $a_{1..m}$.

&emsp; &emsp; &emsp; Thus $a_{1..m}$ satisfies the theorem, which gives a contradiction.

&emsp; &emsp; Therefore $\sum_{k=1}^p a_k &gt; 0$.

&emsp; &emsp; Similarly $\sum_{k=p}^m a_k &gt; 0$.

&emsp; Therefore $a_{1..m}$ has initial and terminal segments all with positive sum.

&emsp; Thus $a_{1..m}$ satisfies the theorem, which gives a contradiction.

Therefore $T(n)$ is true for any $n \in \nn$.
If the sum  is positive, at least one element is positive. Pick that one as "sequence" (of length 1).