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If a_n=(a_{n-1})(a_{n-3})-(a_{n-4}), a_1=1, a_2=2, a_3=3 and a_4=4
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19 Apr 2022, 01:00
19 Apr 2022, 01:00
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Question Stats:
91% (01:38) correct
8% (01:29) wrong based on 12 sessions
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If \(a_n=(a_{n-1})(a_{n-3})-(a_{n-4})\), \(a_1=1\), \(a_2=2\), \(a_3=3\) and \(a_4=4\), what is the value of \(a_6\)?
A. 7
B. 11
C. 19
D. 42
E. 130
A. 7
B. 11
C. 19
D. 42
E. 130
Part of the project: Skill Builder Project for a Q170
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GRE Math Essentials for a TOP Quant Score - Q170!!! (2021)
GRE Geometry Formulas for a Q170
GRE - Math Book
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials for a TOP Quant Score - Q170!!! (2021)
GRE Geometry Formulas for a Q170
GRE - Math Book
Re: If a_n=(a_{n-1})(a_{n-3})-(a_{n-4}), a_1=1, a_2=2, a_3=3 and a_4=4
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19 Apr 2022, 03:05
19 Apr 2022, 03:05
1
\(a_n=(a_{n-1})(a_{n-3})-(a_{n-4})\)
We are given that \(a_1=1\), \(a_2=2\), \(a_3=3\) and \(a_4=4\)
\(a_5\) = \(a_4\) * \(a_2\) - \(a_1\)
\(a_5\) \(= 4 * 2 - 1 = 7\)
Now,
\(a_6\) = \(a_5\) * \(a_3\) - \(a_2\)
\(a_6\) \(= 7 * 3 - 2 = 19\)
Answer C
We are given that \(a_1=1\), \(a_2=2\), \(a_3=3\) and \(a_4=4\)
\(a_5\) = \(a_4\) * \(a_2\) - \(a_1\)
\(a_5\) \(= 4 * 2 - 1 = 7\)
Now,
\(a_6\) = \(a_5\) * \(a_3\) - \(a_2\)
\(a_6\) \(= 7 * 3 - 2 = 19\)
Answer C
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Re: If a_n=(a_{n-1})(a_{n-3})-(a_{n-4}), a_1=1, a_2=2, a_3=3 and a_4=4
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19 Apr 2022, 11:25
19 Apr 2022, 11:25
1
Carcass wrote:
If \(a_n=(a_{n-1})(a_{n-3})-(a_{n-4})\), \(a_1=1\), \(a_2=2\), \(a_3=3\) and \(a_4=4\), what is the value of \(a_6\)?
A. 7
B. 11
C. 19
D. 42
E. 130
A. 7
B. 11
C. 19
D. 42
E. 130
Given: \(a_n=(a_{n-1})(a_{n-3})-(a_{n-4})\)
So, \(a_5=(a_{5-1})(a_{5-3})-(a_{5-4})\)
\(=(a_{4})(a_{2})-(a_{1})\)
\(=(4)(2)-1\)
\(=7\)
So, \(a_5=7\)
Next, we can write: \(a_6=(a_{6-1})(a_{6-3})-(a_{6-4})\)
\(=(a_{5})(a_{3})-(a_{2})\)
\(=(7)(3)-(2)\)
\(=19\)
Answer: C
