Let \(a_1,a_2,...a_{100}\) be real numbers each less than \(1\),which satisfy, \[a_1+a_2+.....a_{100} > 1\]

\(1.\) Let \(n_0\) be the smallest integer \(n\) such that \[a_1+a_2+.....a_n > 1\] Show that the sums \(a_{n_0},a_{n_0}+a_{n_0-1} ,......,a_{n_0}+.....+a_1\) are positive

\(2.\)Show that there exists two integers \(p\) and \(q\),\(p < q\),such that the numbers \[a_q,a_q+a_{q-1},....,a_q+.....+a_p\] and \[a_p,a_p+a_{p+1},....,a_p+.....+a_q\]

are all positive

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TopNewestFor 1., we prove by contradiction. Suppose one the sums, WLOG \( (a_{n_0} + \dots + a_{n_0 - x}) \) is not positive., i.e.\( (\leq 0) \)

Now, As, \( a_1 + a_2 + \dots + a_{n_0} > 1 \)

\( \Rightarrow (a_1 + a_2 + \dots + a_{n_0 - (x+1)}) + ( a_{n_0 - x} + \dots + a_{n_0}) > 1 \)

\( \Rightarrow (a_1 + a_2 + \dots + a_{n_0 - (x+1)}) > 1 - ( a_{n_0 - x} + \dots + a_{n_0}) \)

\( \Rightarrow (a_1 + a_2 + \dots + a_{n_0 - (x+1)}) > 1 \) (Since \( ( a_{n_0 - x} + \dots + a_{n_0}) \) is not positive)

But this contradicts the fact \( a_{n_0} \) is the smallest n for which \( a_1 + a_2 + \dots + a_n > 1 \).

Therefore our supposition is wrong and no sum is not positive, i.e. all the sums are positive – Siddhartha Srivastava · 3 years, 3 months ago

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