GMAT Club World Cup 2022 (DAY 5): The infinite decreasing sequence

GMAT CLUB Official Explanation:

The infinite sequence \(a_1\), \(a_2\), …, \(a_n\), … is such that \(a_{n} < a_{n-1}\), for all \(n > 1\). Does the sequence have more than 12 positive numbers ?

What does \(a_{n} < a_{n-1}\) tell us? Well, this means that every successive term (\(a_n\)) is less than the previous term (\(a_{n-1}\)). So, the sequence at hand is decreasing. (1) \(\frac{a_{14}}{a_{13}}=\frac{\sqrt{10}}{\pi}\).

What is all this about?

Firs of all, since the sequence is decreasing, then we know that \(a_{14} < a_{13}\). Now, IF \(a_{13}\) were positive, then after dividing \(a_{14} < a_{13}\) by \(a_{13}\) we'd have that \(\frac{a_{14}}{a_{13}} < 1\). But since \(\sqrt{10} > \pi\) (\(^*\) check below as to why?), then \(\frac{\sqrt{10}}{\pi} > 1\) and thus \(\frac{a_{14}}{a_{13}} >1\).

Therefore, our assumption that \(a_{13}\) is positive was wrong and in reality, \(a_{13} < 0\). But since \(a_{14} < a_{13}\), then \(a_{14} < a_{13} < 0\). The question asks whether the sequence has more than 12 positive numbers, and \(a_{14} < a_{13} < 0\) means that the sequence can have at most 12 positive numbers, NOT more than 12. We have a definite NO answer to the question.

Sufficient.

(2) The product of any two of the first 14 terms is positive.

Well, the above would be true if ALL first 14 terms were positive (answer YES) as well as if ALL first 14 terms were negative (answer NO). So, this statement is not sufficient.


Answer: A