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Question Stats:
37% (02:24) correct
63%
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wrong
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The infinite sequence \(a_1\), \(a_2\),…, \(a_n\), … is such that \(a_{n} = a_{n-1} + a_{n - 2}\), for all n > 2. What is the value of \(a_6\) ?
(1) The average (arithmetic mean) of \(a_4\) and \(a_7 \) is 0.
(2) \(\frac{a_7}{a_5}= 1\)
(1) The average (arithmetic mean) of \(a_4\) and \(a_7 \) is 0.
(2) \(\frac{a_7}{a_5}= 1\)
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Updated on: 15 Nov 2022, 00:10
Originally posted by gmatophobia on 14 Nov 2022, 06:05.
Last edited by gmatophobia on 15 Nov 2022, 00:10, edited 1 time in total.
Last edited by gmatophobia on 15 Nov 2022, 00:10, edited 1 time in total.
Edited the solution as per latest version of the question.
Kudos
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\(a_n\) is related as the sum of the previous two terms.
We can start with statement 2 as its a has an multiplicative relationship.
Statement 2
\(\frac{a_7}{a_5}\) = 1
We can represent \(a_7\) in terms of \(a_5\) and can find the value of a_6.
This statement is sufficient and we can eliminate A , C and E.
Statement 1
As the numbers are related to each other, we use this relationship to obtain the value of \(a_6\)
The detailed working is attached.
IMO D
We can start with statement 2 as its a has an multiplicative relationship.
Statement 2
\(\frac{a_7}{a_5}\) = 1
We can represent \(a_7\) in terms of \(a_5\) and can find the value of a_6.
This statement is sufficient and we can eliminate A , C and E.
Statement 1
As the numbers are related to each other, we use this relationship to obtain the value of \(a_6\)
The detailed working is attached.
IMO D
Attachments
Screenshot 2022-11-15 123849.png [ 165.99 KiB | Viewed 2160 times ]
14 Nov 2022, 08:14
Kudos
Bookmarks
The infinite sequence \(a_1\), \(a_2\),…, \(a_n\), … is such that \(a_{n} = a_{n-1} + a_{n - 2}\), for all n > 2. What is the value of \(a_6\) ?
(1) The average (arithmetic mean) of \(a_4\) and \(a_7 \) is 0.
\(a_{n} = a_{n-1} + a_{n - 2}\)
\(a_{7} = a_{6} + a_{5}\)
Thus, \(a_7+a_4=a_6+a_5+a_4=a_6+a_6=2a_6=0……….a_6=0\)
Sufficient
(2) \(\frac{a_7}{a_5}= 1\)
\(a_7=a_5\)
\(a_{7} = a_{6} + a_{5}=a_5………a_6=0\)
Sufficient
D
(1) The average (arithmetic mean) of \(a_4\) and \(a_7 \) is 0.
\(a_{n} = a_{n-1} + a_{n - 2}\)
\(a_{7} = a_{6} + a_{5}\)
Thus, \(a_7+a_4=a_6+a_5+a_4=a_6+a_6=2a_6=0……….a_6=0\)
Sufficient
(2) \(\frac{a_7}{a_5}= 1\)
\(a_7=a_5\)
\(a_{7} = a_{6} + a_{5}=a_5………a_6=0\)
Sufficient
D
