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Bunuel
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My answer is E.

T3 = -1*3 - (2*-2)= 1

T4=3*1 - (2*-2) = 5

I'll be happy if I got this right, I am terrible at sequences!~!
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Bunuel
In a certain sequence, the term tn is defined as \(t_n = 3t_{n-1} - 2t_{n-2}\) for all n > 2. If \(t_1 = -2\) and \(t_2 = -1\) , then \(t_4 =\)

A. -10
B. -8
C. -3
D. 1
E. 5


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T3 = 3(t2) - 2(t1) = 3(-1) - 2(-2) = 1
T4 = 3(t3) - 2(t2) = 3(1) - 2(-1) = 5

Option E
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t3=3t2-2t1 = 1 (putting the value of t1 & t2, we can find out value of t3)
t4=3t3-2t2 = 5 (put the value of t2 &t3 in this and find out t4)

Hence answer is E

Thanks,
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Bunuel
In a certain sequence, the term tn is defined as \(t_n = 3t_{n-1} - 2t_{n-2}\) for all n > 2. If \(t_1 = -2\) and \(t_2 = -1\) , then \(t_4 =\)

A. -10
B. -8
C. -3
D. 1
E. 5


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\(t_3 = 3t_2 - 2t_1=3 *(-1) - 2*(-2)=1\)
\(t_4 = 3t_3 - 2t_2=3 *(1) - 2*(-1)=5\)

Answer is E
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Bunuel
In a certain sequence, the term tn is defined as \(t_n = 3t_{n-1} - 2t_{n-2}\) for all n > 2. If \(t_1 = -2\) and \(t_2 = -1\) , then \(t_4 =\)

A. -10
B. -8
C. -3
D. 1
E. 5


Kudos for a correct solution.

t4=3t3-2t2
t3=3t2-2t1=-3-2(-2)=1
Therefore, t4=3*1-2*-1=5. Asnwer=E
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Bunuel
In a certain sequence, the term tn is defined as \(t_n = 3t_{n-1} - 2t_{n-2}\) for all n > 2. If \(t_1 = -2\) and \(t_2 = -1\) , then \(t_4 =\)

A. -10
B. -8
C. -3
D. 1
E. 5


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MAGOOSH OFFICIAL SOLUTION:
Attachment:
Arithmetic_Recursion.png
Arithmetic_Recursion.png [ 18.11 KiB | Viewed 12982 times ]
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Bunuel
In a certain sequence, the term tn is defined as \(t_n = 3t_{n-1} - 2t_{n-2}\) for all n > 2. If \(t_1 = -2\) and \(t_2 = -1\) , then \(t_4 =\)

A. -10
B. -8
C. -3
D. 1
E. 5


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Given: term_(n) = 3term_(n-1) - 2term_(n-2)

So, for example, term_3 = 3term_(3-1) - 2term_(3-2) = 3term_2 - 2term_1 = 3(-1) - 2(-2) = 1

Similarly, term_4 = 3term_3 - 2term_2 = 3(1) - 2(-1) = 5

Answer: E
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