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16 Sep 2014, 00:54
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D
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The sequence of 10 numbers \(a_1\), \(a_2\), ..., \(a_{10}\) is defined such that \(a_{n}=\frac{a_{n-1}}{n}\) for all \(1 < n \leq 10\). How many terms of the sequence are greater than \(\frac{1}{2}\)?
(1) \(a_2 = 5\)
(2) \(a_1 - a_2 = 5\)
(1) \(a_2 = 5\)
(2) \(a_1 - a_2 = 5\)
16 Sep 2014, 00:54
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Official Solution:
The sequence of 10 numbers \(a_1\), \(a_2\), ..., \(a_{10}\) is defined such that \(a_{n}=\frac{a_{n-1}}{n}\) for all \(1 < n \leq 10\). How many terms of the sequence are greater than \(\frac{1}{2}\)?
We are given a sequence of numbers defined by a formula. For instance: \(a_{2}=\frac{a_{1}}{2}\), \(a_{3}=\frac{a_{2}}{3}\), \(a_{4}=\frac{a_{3}}{4}\), and so on. The question asks how many terms in the sequence are greater than \(\frac{1}{2}\). Observe that if we know any term of the sequence, we can determine all the other terms and thus answer the question.
(1) \(a_2=5\).
As mentioned earlier, knowing any term is sufficient to find all other terms. For example, \(a_{2}=\frac{a_{1}}{2}=5\), which gives \(a_1=10\). Then, \(a_{3}=\frac{a_{2}}{3}=\frac{5}{3}\), and so on. Sufficient.
(2) \(a_1-a_2=5\).
Substituting \(a_{2}=\frac{a_{1}}{2}\) into the above, gives \(a_1-\frac{a_{1}}{2}=5\). We can solve for \(a_1\), which results in the same situation as knowing one term. Sufficient.
Answer: D
The sequence of 10 numbers \(a_1\), \(a_2\), ..., \(a_{10}\) is defined such that \(a_{n}=\frac{a_{n-1}}{n}\) for all \(1 < n \leq 10\). How many terms of the sequence are greater than \(\frac{1}{2}\)?
We are given a sequence of numbers defined by a formula. For instance: \(a_{2}=\frac{a_{1}}{2}\), \(a_{3}=\frac{a_{2}}{3}\), \(a_{4}=\frac{a_{3}}{4}\), and so on. The question asks how many terms in the sequence are greater than \(\frac{1}{2}\). Observe that if we know any term of the sequence, we can determine all the other terms and thus answer the question.
(1) \(a_2=5\).
As mentioned earlier, knowing any term is sufficient to find all other terms. For example, \(a_{2}=\frac{a_{1}}{2}=5\), which gives \(a_1=10\). Then, \(a_{3}=\frac{a_{2}}{3}=\frac{5}{3}\), and so on. Sufficient.
(2) \(a_1-a_2=5\).
Substituting \(a_{2}=\frac{a_{1}}{2}\) into the above, gives \(a_1-\frac{a_{1}}{2}=5\). We can solve for \(a_1\), which results in the same situation as knowing one term. Sufficient.
Answer: D
28 Jun 2020, 15:10
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How is statement 2 sufficient?
if a1 - a2 = 5,
solving we get: 2a1 - a1 = 10
a1 = 10.
If we plug a1 back in the expression we get:
a1 = \(\frac{a1-a1}{2}\) is equal to 10, but what do we do from here?
if a1 - a2 = 5,
solving we get: 2a1 - a1 = 10
a1 = 10.
If we plug a1 back in the expression we get:
a1 = \(\frac{a1-a1}{2}\) is equal to 10, but what do we do from here?
