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16 Sep 2014, 00:59
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The sequence \(a_1\), \(a_2\), \(a_3\), ..., \(a_n\), ... is such that \(a_{n}-a_{n+1}=a_{n+1}\) for all integers \(n \ge 1\). What is the value of the fourth term in this sequence?
(1) The first term of the sequence is 1
(2) The third term of the sequence is \(\frac{1}{4}\)
(1) The first term of the sequence is 1
(2) The third term of the sequence is \(\frac{1}{4}\)
16 Sep 2014, 00:59
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Official Solution:
The sequence \(a_1\), \(a_2\), \(a_3\), ..., \(a_n\), ... is such that \(a_{n}-a_{n+1}=a_{n+1}\) for all integers \(n \ge 1\). What is the value of the fourth term in this sequence?
From \(a_{n} - a_{n+1} = a_{n+1}\), we deduce that \(a_{n+1} = \frac{a_{n}}{2}\).
(1) The first term of the sequence is 1.
Given \(a_1 = 1\), and using the relationship \(a_{n+1} = \frac{a_{n}}{2}\), we get: \(a_2 = \frac{1}{2}\), \(a_3 = \frac{1}{4}\), and \(a_4 = \frac{1}{8}\). Sufficient.
(2) The third term of the sequence is \(\frac{1}{4}\).
Given \(a_3 = \frac{1}{4}\), we find \(a_4 = \frac{a_3}{2} = \frac{1}{8}\). Sufficient.
Answer: D
The sequence \(a_1\), \(a_2\), \(a_3\), ..., \(a_n\), ... is such that \(a_{n}-a_{n+1}=a_{n+1}\) for all integers \(n \ge 1\). What is the value of the fourth term in this sequence?
From \(a_{n} - a_{n+1} = a_{n+1}\), we deduce that \(a_{n+1} = \frac{a_{n}}{2}\).
(1) The first term of the sequence is 1.
Given \(a_1 = 1\), and using the relationship \(a_{n+1} = \frac{a_{n}}{2}\), we get: \(a_2 = \frac{1}{2}\), \(a_3 = \frac{1}{4}\), and \(a_4 = \frac{1}{8}\). Sufficient.
(2) The third term of the sequence is \(\frac{1}{4}\).
Given \(a_3 = \frac{1}{4}\), we find \(a_4 = \frac{a_3}{2} = \frac{1}{8}\). Sufficient.
Answer: D
19 Oct 2023, 00:55
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
