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18 Jun 2008, 10:23
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D
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Difficulty:
55%
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Question Stats:
61% (01:43) correct
39%
(01:55)
wrong
based on 823
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In the sequence of nonzero numbers \(t_1\), \(t_2\), \(t_3\), …, \(t_n\), …, \(t_{n+1}=\frac{t_n}{2}\) for all positive integers n. What is the value of \(t_5\)?
(1) \(t_3 = \frac{1}{4}\)
(2) \(t_1 - t_5 = \frac{15}{16}\)
(1) \(t_3 = \frac{1}{4}\)
(2) \(t_1 - t_5 = \frac{15}{16}\)
18 Jun 2008, 13:43
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quantum
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19 Dec 2010, 16:42
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gettinit
In the sequence of nonzero numbers t1, t2, t3, …, tn, …, tn+1 = tn / 2 for all positive integers n. What is the value of t5?
Given: \(t_{n+1}=\frac{t_n}{2}\). So \(t_2=\frac{t_1}{2}\), \(t_3=\frac{t_2}{2}=\frac{t_1}{4}\), \(t_4=\frac{t_3}{2}=\frac{t_1}{8}\), ...
Basically we have geometric progression with common ratio \(\frac{1}{2}\): \(t_1\), \(\frac{t_1}{2}\), \(\frac{t_1}{4}\), \(\frac{t_1}{8}\), ... --> \(t_n=\frac{t_1}{2^{n-1}}\).
Question: \(t_5=\frac{t_1}{2^4}=?\)
(1) \(t_3=\frac{1}{4}\) --> we can get \(t_1\) --> we can get \(t_5\). Sufficient.
(2) \(t_1-t_5=2^4*t_5-t_5=\frac{15}{16}\) --> we can get \(t_5\). Sufficient.
Answer: D.
Generally for arithmetic (or geometric) progression if you know:
- any particular two terms,
- any particular term and common difference (common ratio),
- the sum of the sequence and either any term or common difference (common ratio),
then you will be able to calculate any missing value of given sequence.
Hope it helps.
