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A series is given by a_n= (a_{n-1})^2

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A series is given by a_n= (a_{n-1})^2 [#permalink] New post   24 Sep 2021, 21:34
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A series is given by \(a_n= (a_{n-1})^2\) with \(a_0 = 3\).

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
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B
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Re: A series is given by a_n= (a_{n-1})^2 [#permalink] New post   25 Sep 2021, 09:13
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\(a_n= (a_{n-1})^2\) & \(a_0 = 3\)

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
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Re: A series is given by a_n= (a_{n-1})^2 [#permalink]
25 Sep 2021, 09:13
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