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A series is given by a_n= (a_{n-1})^2
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24 Sep 2021, 21:34
24 Sep 2021, 21:34
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A series is given by \(a_n= (a_{n-1})^2\) with \(a_0 = 3\).
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
\(a_5-a_4\) |
\(a_6-a_5\) |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Re: A series is given by a_n= (a_{n-1})^2
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25 Sep 2021, 09:13
25 Sep 2021, 09:13
1
\(a_n= (a_{n-1})^2\) & \(a_0 = 3\)
Add \(a_5\) to both the quantity.
Qt A \(= -a_4\)
Qt B \(= a_6\)
As \(a_n\) is a square, the value will always be \(> 0\)
Qt B > Qt A
Answer B
Add \(a_5\) to both the quantity.
Qt A \(= -a_4\)
Qt B \(= a_6\)
As \(a_n\) is a square, the value will always be \(> 0\)
Qt B > Qt A
Answer B
gmatclubot
Re: A series is given by a_n= (a_{n-1})^2 [#permalink]
25 Sep 2021, 09:13
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