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For integer n ≥ 3, a sequence is defined as an = (an – 1)2
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17 Sep 2017, 02:36
17 Sep 2017, 02:36
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38% (04:21) wrong based on 42 sessions
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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
Re: For integer n ≥ 3, a sequence is defined as an = (an – 1)2
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10 Apr 2020, 14:11
10 Apr 2020, 14:11
Expert Reply
\(a_4 = a_1(a_2+a_3)\)
\(a_1\) = 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
\(a_1\) = 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
Re: For integer n ≥ 3, a sequence is defined as an = (an – 1)2
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11 Apr 2020, 01:47
11 Apr 2020, 01:47
1
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
\(a_1\) = 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 ?
