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If the sequence a1, a2, a3, ..., an is such that a1 = 6 and
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18 Mar 2019, 09:20
18 Mar 2019, 09:20
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Question Stats:
92% (01:03) correct
7% (00:00) wrong based on 27 sessions
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If the sequence \(a_1\),\(a_2\), \(a_3\), ..., \(a_n\) is such that \(a_1 = 6\) and \(a_n = 2a_{(n - 1)}\) for \(n > 1\), then \(a_6 - a_5 =\)
A. 12
B. 48
C. 64
D. 96
E. 192
A. 12
B. 48
C. 64
D. 96
E. 192
Re: If the sequence a1, a2, a3, ..., an is such that a1 = 6 and
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19 Mar 2019, 05:43
19 Mar 2019, 05:43
1
a1=6
an=2a(n-1)
a2=2a(2-1)
a2=2(6)=12
a3=2(12)=24
a5=96 a6=192
a6-a5= 96
an=2a(n-1)
a2=2a(2-1)
a2=2(6)=12
a3=2(12)=24
a5=96 a6=192
a6-a5= 96
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If the sequence a1, a2, a3, ..., an is such that a1 = 6 and
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25 Oct 2022, 10:53
25 Oct 2022, 10:53
Given that \(a_1 = 6\) and \(a_n = 2a_{(n - 1)}\) for \(n > 1\) and we need to find the value of \(a_6 - a_5 =\)
\(a_1 = 6\) and \(a_n = 2a_{(n - 1)}\)
=> \(a_2 = 2*a_1\) = 2 * 6 = \(2^1\) * 6 = \(2^{(2 - 1)}\) * 6
=> \(a_3 = 2*a_2\) = 2 * \(2^1\) * 6 = \(2^2\) * 6 = \(2^{(3 - 1)}\) * 6
Looking at the pattern
=> \(a_5 = 2^{(5 - 1)}\) * 6 = \(2^4\) * 6
=> \(a_6 = 2^{(6 - 1)}\) * 6 = \(2^5\) * 6
=> \(a_6 - a_5 =\) = \(2^5\) * 6 - \(2^4\) * 6 = \(2^4\) * 6 *(2-1) = \(2^4\) * 6 = 16*6 = 96
So, Answer will be D
Hope it helps!
Watch the following video to learn How to Sequence problems
\(a_1 = 6\) and \(a_n = 2a_{(n - 1)}\)
=> \(a_2 = 2*a_1\) = 2 * 6 = \(2^1\) * 6 = \(2^{(2 - 1)}\) * 6
=> \(a_3 = 2*a_2\) = 2 * \(2^1\) * 6 = \(2^2\) * 6 = \(2^{(3 - 1)}\) * 6
Looking at the pattern
=> \(a_5 = 2^{(5 - 1)}\) * 6 = \(2^4\) * 6
=> \(a_6 = 2^{(6 - 1)}\) * 6 = \(2^5\) * 6
=> \(a_6 - a_5 =\) = \(2^5\) * 6 - \(2^4\) * 6 = \(2^4\) * 6 *(2-1) = \(2^4\) * 6 = 16*6 = 96
So, Answer will be D
Hope it helps!
Watch the following video to learn How to Sequence problems
