GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
For integer n ≥ 3, a sequence is defined as an = (an – 1)2
[#permalink]
17 Sep 2017, 02:36
17 Sep 2017, 02:36
Expert Reply
3
Bookmarks
00:00
Question Stats:
59% (03:07) correct
40% (04:26) wrong based on 37 sessions
Hide Show timer Statistics
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
[#permalink]
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
[#permalink]
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 ?
