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25 Oct 2022, 00:15
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C
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Difficulty:
25%
(medium)
Question Stats:
78% (02:17) correct
22%
(02:39)
wrong
based on 50
sessions
History
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The sequence \(a_1, \ a_2, \ ..., \ a_n, \ ...\) is such that that \(a_1 = 2\), \(a_2 = 3\) and \(a_n = a_{n-1}*a_{n-2}\), for n > 2. What is the value of \(a_8\) ?
A. \(2^7*3^{14}\)
B. \(2^7*3^{13}\)
C. \(2^8*3^{13}\)
D. \(2^8*3^{14}\)
E. \(2^9*3^{13}\)
A. \(2^7*3^{14}\)
B. \(2^7*3^{13}\)
C. \(2^8*3^{13}\)
D. \(2^8*3^{14}\)
E. \(2^9*3^{13}\)
25 Oct 2022, 11:43
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Given that \(a_1 = 2\), \(a_2 = 3\) and \(a_n = a_{n-1}*a_{n-2}\), for n > 2 and we need to find the value of \(a_8\)
\(a_1 = 2\), \(a_2 = 3\) and \(a_n = a_{n-1}*a_{n-2}\)
=> \(a_3 = a_2*a_1\) = \(3^1\) * \(2^1\)
=> \(a_4 = a_3*a_2\) = 3*2 * 3 = \(3^2\) * \(2^1\)
=> \(a_5 = a_4*a_3\) = \(3^2\) * \(2^1\) * \(3^1\) * \(2^1\) = \(3^3\) * \(2^2\)
=> \(a_6 = a_5*a_4\) = \(3^3\) * \(2^2\) * \(3^2\) * \(2^1\) = \(3^5\) * \(2^3\)
=> \(a_7 = a_6*a_5\) = \(3^5\) * \(2^3\) * \(3^3\) * \(2^2\) = \(3^8\) * \(2^5\)
=> \(a_8 = a_7*a_6\) = \(3^8\) * \(2^5\) * \(3^5\) * \(2^3\) = \(3^{13}\) * \(2^8\)
So, Answer will be C
Hope it helps!
Watch the following video to learn How to Sequence problems
\(a_1 = 2\), \(a_2 = 3\) and \(a_n = a_{n-1}*a_{n-2}\)
=> \(a_3 = a_2*a_1\) = \(3^1\) * \(2^1\)
=> \(a_4 = a_3*a_2\) = 3*2 * 3 = \(3^2\) * \(2^1\)
=> \(a_5 = a_4*a_3\) = \(3^2\) * \(2^1\) * \(3^1\) * \(2^1\) = \(3^3\) * \(2^2\)
=> \(a_6 = a_5*a_4\) = \(3^3\) * \(2^2\) * \(3^2\) * \(2^1\) = \(3^5\) * \(2^3\)
=> \(a_7 = a_6*a_5\) = \(3^5\) * \(2^3\) * \(3^3\) * \(2^2\) = \(3^8\) * \(2^5\)
=> \(a_8 = a_7*a_6\) = \(3^8\) * \(2^5\) * \(3^5\) * \(2^3\) = \(3^{13}\) * \(2^8\)
So, Answer will be C
Hope it helps!
Watch the following video to learn How to Sequence problems
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