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A sequence has terms a1 , a2 , a3 ,., an , where a1 = 2 and ge
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20 May 2024, 23:22
20 May 2024, 23:22
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53% (02:04) correct
46% (02:18) wrong based on 28 sessions
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A sequence has terms \(a_1, a_2, a_3,……., a_n\), where \(a_1 = 2\) and general term is given as \(a_n=a_{(n-1)} \times \frac{1}{2}\) .
Quantity A |
Quantity B |
\(a_6\) |
\(2^{14}(a_{20})\) |
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A sequence has terms a1 , a2 , a3 ,., an , where a1 = 2 and ge
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21 May 2024, 19:15
21 May 2024, 19:15
3
First, let's calculate a few terms to identify a pattern.
\(a_2 = a_1 * \frac{1}{2} = 2 * \frac{1}{2} = 1 = 2^0\)
\(a_3 = a_2 * \frac{1}{2} = 2 * \frac{1}{2} * \frac{1}{2} = \frac{1}{2} = 2^{-1}\)
We can see that \(a_n\) can also be expressed as \(a_n = \frac{1}{2^{n-1}}\).
Therefore, Quantity A = \(a_6 = \frac{1}{2^{6-1}} = \frac{1}{2^5}\)
Quantity B can be simplified:
\(2^{14} (a_{20}) = 2^{14}(\frac{1}{2^{20-1}}) = 2^{14}(\frac{1}{2^{19}}) = \frac{1}{2^{19-14}} = \frac{1}{2^5}\)
Quantity A and B are equal.
\(a_2 = a_1 * \frac{1}{2} = 2 * \frac{1}{2} = 1 = 2^0\)
\(a_3 = a_2 * \frac{1}{2} = 2 * \frac{1}{2} * \frac{1}{2} = \frac{1}{2} = 2^{-1}\)
We can see that \(a_n\) can also be expressed as \(a_n = \frac{1}{2^{n-1}}\).
Therefore, Quantity A = \(a_6 = \frac{1}{2^{6-1}} = \frac{1}{2^5}\)
Quantity B can be simplified:
\(2^{14} (a_{20}) = 2^{14}(\frac{1}{2^{20-1}}) = 2^{14}(\frac{1}{2^{19}}) = \frac{1}{2^{19-14}} = \frac{1}{2^5}\)
Quantity A and B are equal.
A sequence has terms a1 , a2 , a3 ,., an , where a1 = 2 and ge
[#permalink]
25 Sep 2024, 18:23
25 Sep 2024, 18:23
nurirachel wrote:
First, let's calculate a few terms to identify a pattern.
\(a_2 = a_1 * \frac{1}{2} = 2 * \frac{1}{2} = 1 = 2^0\)
\(a_3 = a_2 * \frac{1}{2} = 2 * \frac{1}{2} * \frac{1}{2} = \frac{1}{2} = 2^{-1}\)
We can see that \(a_n\) can also be expressed as \(a_n = \frac{1}{2^{n-1}}\).
Therefore, Quantity A = \(a_6 = \frac{1}{2^{6-1}} = \frac{1}{2^5}\)
Quantity B can be simplified:
\(2^{14} (a_{20}) = 2^{14}(\frac{1}{2^{20-1}}) = 2^{14}(\frac{1}{2^{19}}) = \frac{1}{2^{19-14}} = \frac{1}{2^5}\)
Quantity A and B are equal.
\(a_2 = a_1 * \frac{1}{2} = 2 * \frac{1}{2} = 1 = 2^0\)
\(a_3 = a_2 * \frac{1}{2} = 2 * \frac{1}{2} * \frac{1}{2} = \frac{1}{2} = 2^{-1}\)
We can see that \(a_n\) can also be expressed as \(a_n = \frac{1}{2^{n-1}}\).
Therefore, Quantity A = \(a_6 = \frac{1}{2^{6-1}} = \frac{1}{2^5}\)
Quantity B can be simplified:
\(2^{14} (a_{20}) = 2^{14}(\frac{1}{2^{20-1}}) = 2^{14}(\frac{1}{2^{19}}) = \frac{1}{2^{19-14}} = \frac{1}{2^5}\)
Quantity A and B are equal.
Doesn't change the answer to the question, but I think the modified sequence is actually:
\(a_n = \frac{1}{2^{n-2}}\).
The way you have it,
\(a_2 = 1/2\) when it should be 1, as you note earlier in your response.
