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

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by BTGmoderatorLU » Wed Jul 31, 2019 4:44 pm

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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
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Source: — Problem Solving |

by Jay@ManhattanReview » Thu Aug 01, 2019 2:40 am
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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by Scott@TargetTestPrep » Sun Aug 04, 2019 10:27 am
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

Scott Woodbury-Stewart
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