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For integer n ≥ 3, a sequence is defined as an = (an – 1)2

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For integer n ≥ 3, a sequence is defined as an = (an – 1)2 [#permalink] New post   17 Sep 2017, 02:36
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For integer n ≥ 3, a sequence is defined as \(a_n = (a_{n - 1})^2 - (a_{n - 2})^2\) and \(a_n > 0\) for all positive integers n. The first term \(a_1\) is 2, and the fourth term is equal to the first term multiplied by the sum of the second and third terms. What is the third term, \(a_3\)?


(A) 0

(B) 3

(C) 5

(D) 10

(E) 16
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C
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Re: For integer n ≥ 3, a sequence is defined as an = (an – 1)2 [#permalink] New post   10 Apr 2020, 14:11
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\(a_4 = a_1(a_2+a_3)\)
\(a_1\) = 2
\(a_3\) = ????

\(a_4\) = \(a_3\)^2 - \(a_2\)^2
2(\(a_2+a_3\)) = \((a_3-a_2)*(a_3+a_2)\)
as \(a_n>0\)
\(a_3+a_2\) can not be zero

\(a_3-a_2 = 2\) ---------------1

\(a_3 = a_2^2 - a_1^2\)
\((a_3 - a_2) * (a_3 + a_2) = -4\)
\((a_3 + a_2) = -2\) ------------2

from 1 and 2
\(a_3 = 0\)

ans c
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Re: For integer n ≥ 3, a sequence is defined as an = (an – 1)2 [#permalink] New post   11 Apr 2020, 01:47
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Carcass wrote:
\(a_4 = a_1(a_2+a_3)\)
\(a_1\) = 2
\(a_3\) = ????

\(a_4\) = \(a_3\)^2 - \(a_2\)^2
2(\(a_2+a_3\)) = \((a_3-a_2)*(a_3+a_2)\)
as \(a_n>0\)
\(a_3+a_2\) can not be zero

\(a_3-a_2 = 2\) ---------------1

\(a_3 = a_2^2 - a_1^2\)
\((a_3 - a_2) * (a_3 + a_2) = -4\)
\((a_3 + a_2) = -2\) ------------2

from 1 and 2
\(a_3 = 0\)

ans c


Option C is 5 and not 0, also, if an > 0, then why will a3 = 0 ?
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Re: For integer n ≥ 3, a sequence is defined as an = (an – 1)2 [#permalink] New post   11 Apr 2020, 03:33
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\(a_4= (a_3)^2- (a_2)^2\)
\(a_4= [a_3- a_2][a_3+a_2]\).....(1)

Also, \(a_4= a_1(a_3+a_2)\).....(2)
From (1) and (2)
\(a_1=a_3-a_2\)
\(a_3=a_1+a_2\)......(3)

\(a_3= (a_2)^2-(a_1)^2\)
\(a_3= [a_2- a_1][a_2+a_1]\)......(4)
From (3) and (4)
\(a_2=a_1+1\)
\(a_2=3\)

\(a_3=a_2+a_1\)=3+2=5
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Re: For integer n ≥ 3, a sequence is defined as an = (an – 1)2 [#permalink] New post   11 Apr 2020, 03:34
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My mistake. Really sorry about

Fixed.

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Re: For integer n ≥ 3, a sequence is defined as an = (an – 1)2 [#permalink] New post   22 Jun 2021, 08:43
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The problem gives two ways to calculate the fourth term: (1) the definition of the sequence tells you that a4 = a32 – a22 and (2) the words
indicate that a4 = a1(a2 + a3) = 2(a2 + a3). Setting these two equal gives a32 –a22 = 2(a2 + a3). Factor the left side: (a3 + a2)(a3 – a2) = 2(a2 + a3). Since an> 0 for all possible n’s, (a3 + a2) does not equal 0 and you can divide bothsides by it: a3 – a2 = 2 and a3 = a2 + 2. Using the definition of a3
, you know
a3 = a22 – a12 = a22 – 4. Substituting for a3 yields: a2 + 2 = a22 – 4 and a22 – a2 – 6 = 0. Factor and solve: (a2 – 3)(a2 + 2) = 0; a2 = 3 or –2. an must be positive, so a2 = 3 and a3 = a2 + 2 = 3 + 2 = 5.
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Re: For integer n ≥ 3, a sequence is defined as an = (an – 1)2 [#permalink]
22 Jun 2021, 08:43
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