An infinite sequence \(a_1, a_2, a_3, … a_n\) can be defin

BTGmoderatorLU wrote:Source: Veritas Prep

An infinite sequence \(a_1, a_2, a_3,\cdots , a_n\) can be defined as \(a_n=3n–a_{n−1}\), where \(n > 1\) and \(a_1=1\). Which of the following represents the first 5 terms of the sequence?

\(A. 1, 5, 4, 8, 16\)
\(B. 1, 5, 4, 8, 7\)
\(C. 1, 6, 10, 2, 13\)
\(D. 1, 2, 3, 5, 10\)
\(E. 1, 7, 16, 28, 43\)

The OA is B

Given term = \(a_n=3n–a_{n−1}\), where \(n > 1\) and \(a_1=1\)

\(a_2=1\) (given);
\(a_2=3n–a_{n−1}=3*2-a_1=6-1=5\);
\(a_3=3n–a_{n−1}=3*3-a_2=9-5=4\);
\(a_4=3n–a_{n−1}=3*4-a_3=12-4=8\);
\(a_5=3n–a_{n−1}=3*5-a_4=15-8=7\)

The correct answer: B

Hope this helps!

-Jay
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BTGmoderatorLU wrote:Source: Veritas Prep

An infinite sequence \(a_1, a_2, a_3,\cdots , a_n\) can be defined as \(a_n=3n–a_{n−1}\), where \(n > 1\) and \(a_1=1\). Which of the following represents the first 5 terms of the sequence?

\(A. 1, 5, 4, 8, 16\)
\(B. 1, 5, 4, 8, 7\)
\(C. 1, 6, 10, 2, 13\)
\(D. 1, 2, 3, 5, 10\)
\(E. 1, 7, 16, 28, 43\)

The OA is B

a_1 = 1

a_2 = 3(2) - 1 = 5

a_3 = 3(3) - 5 = 4

a_4 = 3(4) - 4 = 8

a_5 = 3(5) - 8 = 7

Alternate Solution:

Since a_2 = 3(2) - 1 = 5, we eliminate C, D and E. To decide between A and B, it’s enough to consider a_5. Notice that a_4 is 8 since in both answer choices the fourth term is 8.

a_5 = 3(5) - 8 = 15 - 8 = 7.

Answer: B