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

Question

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

(A) 0

(B) 3

(C) 5

(D) 10

(E) 16

EgmatQuantExpert · Answer

Hey  Bunuel ,
can you please confirm whether the given expression for  a_n  is  (a_n - 1)^2 - (a_n - 2)^2  or  (a_{n - 1})^2 - (a_{n - 2})^2

Bunuel · Answer

Edited. Thank you for noticing.

rakeshmishra5690 · Answer

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

LipiSinha · Answer

a4=(a3)sq-(a2)sq
a4=2(a2+a3)

Therefore, (a3)sq-(a2)sq = 2(a2+a3)
 (a3+a2)(a3-a2) = 2(a2+a3)
On simplifying, a3-a2 =2
 a2 = a3-2

And, a3=(a2)sq-(a1)sq
 a3=(a2)sq-4
 a3=(a3-2)sq-4
on simplifying, (a3)sq=5a3

The only option that satisfies this condition in C.

EgmatQuantExpert · Answer

Solution

Given:  
 • For integer n ≥ 3, a sequence is defined as  a_n = (a_{n - 1})^2 - (a_{n - 2})^2 
• Also, for all positive integers n,  a_n > 0 
• The first term  a_1  is 2
• The fourth term is equal to the first term multiplies by the sum of the second and third terms

To find:  
 • The value of the third term  a_3

Approach and Working:   
From the given information, we can write:
 •  a_4 = a_3^2 – a_2^2 = (a_3 + a_2) (a_3 – a_2)

It is also given that:
 •  a_4 = a_1 (a_2 + a_3) = 2 (a_2 + a_3)

Equating the value of  a_4 , obtained from both the equations,
 •  (a_3 + a_2) (a_3 – a_2) = 2 (a_2 + a_3) 
Or,  a_3 – a_2 = 2   a_n > 0 ,  (a_3 + a_2) ≠ 0 ]

We can also say:
 •  a_3 = a_2^2 – a_1^2 = a_2^2 – 4

Replacing the value of a_2 from the previous equation, we can write
 •  a_3 = (a_3 – 2)^2 – 4  
Solving, we get  a_3 = 5   a_3  cannot be equal to 0]

Hence, the correct answer is option C.

Answer: B

nick1816 · Answer

a_4= (a_3)^2- (a_2)^2 
 a_4=    .....(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=    ......(4)
From (3) and (4)
 a_2=a_1+1 
 a_2=3

a_3=a_2+a_1 =3+2=5

bumpbot · Answer

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

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

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