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Sub 505 Level|   Sequences|                                       
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Walkabout
In a certain sequence, the term \(x_n\) is given by the formula \(x_n=2*x_{n-1}-\frac{1}{2}*x_{n-2}\) for all \(n\geq{2}\). If \(x_0=3\) and \(x_1=2\), what is the value of \(x_3\)?

(A) 2.5
(B) 3.125
(C) 4
(D) 5
(E) 6.75


X(3)= 2*X(2)-X(1)/2
X(2)=2X(1)-X(0)/2=2*2-3/2=5/2
X(3)= 2*5/2-2/2=5-1=4 Answer is C
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Walkabout
In a certain sequence, the term \(x_n\) is given by the formula \(x_n=2*x_{n-1}-\frac{1}{2}*x_{n-2}\) for all \(n\geq{2}\). If \(x_0=3\) and \(x_1=2\), what is the value of \(x_3\)?

(A) 2.5
(B) 3.125
(C) 4
(D) 5
(E) 6.75

(Note that the * symbol in both the question stem and in this solution indicates multiplication.)

We are given that X(n) = 2 * X(n-1) – ½ * X(n-2), for all n=>2. This is called a recursive formula, which means that we need to know prior terms before we can compute the subsequent terms. For example, if we want to know X(2), we must know both X(1) and X(0), because X(2) is equal to 2 * X(1) – ½ * X(0).

We are given X(1) = 2 and X(0) = 3. So, when n is 2, X(2) would be calculated as follows:

X(2) = 2 * X(1) – ½ * X(0)

X(2) = 2 * 2 – ½ * 3

X(2) = 4 – 1.5

X(2) = 2.5

Now we are ready to determine the value of X(3). In this case, n = 3, X(1) is 2, and X(2) is 2.5. We plug these values into the recursive formula given in the question stem:

X(3) = 2 * X(2) – ½ * X(1)

X(3) = 2 * 2.5 – ½ * 2

X(3) = 5 – 1

X(3) = 4

Answer is C.
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If you have the OG 2017: No! you are not crazy it actually says the formula is used for all n>=2n. And on the answer choices it says n>= 2.

This made me take more than 10 minutes to solve because it just did not make sense. Cause if its n>= 2n that means that n = negative integer. BUT we don't have negative integers, so.... wthk am I supposed to do?

JUST WANTED TO LET YOU ALL KNOW! HUGEEEEE MISTAKE ON OG.
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Bunuel
Walkabout
In a certain sequence, the term \(x_n\) is given by the formula \(x_n=2*x_{n-1}-\frac{1}{2}*x_{n-2}\) for all \(n\geq{2}\). If \(x_0=3\) and \(x_1=2\), what is the value of \(x_3\)?

(A) 2.5
(B) 3.125
(C) 4
(D) 5
(E) 6.75

We have a formula to calculate the value of the terms in the sequence starting from \(x_2\): \(x_n=2*x_{n-1}-\frac{1}{2}*x_{n-2}\). Hence:

\(x_2=2*x_{1}-\frac{1}{2}*x_{0}=2*2-\frac{1}{2}*3=\frac{5}{2}\);

\(x_3=2*x_{2}-\frac{1}{2}*x_{1}=2*\frac{5}{2}-\frac{1}{2}*2=4\).

Answer: C.

Hope it's clear.

Hello Bunuel, i always get confused with such long and weired formulas.

A few questions: How do you call mathematecally n-1 and n-2 written as subscript ? and another question how did you figure out that n-1 is 2 and n-2 is 3 when plugging in ? thank you :)
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Bunuel
Walkabout
In a certain sequence, the term \(x_n\) is given by the formula \(x_n=2*x_{n-1}-\frac{1}{2}*x_{n-2}\) for all \(n\geq{2}\). If \(x_0=3\) and \(x_1=2\), what is the value of \(x_3\)?

(A) 2.5
(B) 3.125
(C) 4
(D) 5
(E) 6.75

We have a formula to calculate the value of the terms in the sequence starting from \(x_2\): \(x_n=2*x_{n-1}-\frac{1}{2}*x_{n-2}\). Hence:

\(x_2=2*x_{1}-\frac{1}{2}*x_{0}=2*2-\frac{1}{2}*3=\frac{5}{2}\);

\(x_3=2*x_{2}-\frac{1}{2}*x_{1}=2*\frac{5}{2}-\frac{1}{2}*2=4\).

Answer: C.

Hope it's clear.

Hello Bunuel, i always get confused with such long and weired formulas.

A few questions: How do you call mathematecally n-1 and n-2 written as subscript ? and another question how did you figure out that n-1 is 2 and n-2 is 3 when plugging in ? thank you :)

\(x_n=2*x_{n-1}-\frac{1}{2}*x_{n-2}\) is a formula giving nth term in terms of (n-1)st and (n-2)nd terms. For example, it allows to find say, 3rd term if you know 3-1=2nd and 3-2=1st terms.
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Bunuel thanks! A few more questions if you dont mind :)

What does for all n≥2 mean ?
Yes, you marked these in yellow x0=3 and x1=2
x0=3 this expression means that the zero term is 3 and x1=2 means that the first term is 2 correct?
If we need to find 3rd term isn’t it enough to know 2nd term ?
When you solve for x2, where from did you get 3 ?
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Bunuel thanks! A few more questions if you dont mind :)

What does for all n≥2 mean ?
Yes, you marked these in yellow x0=3 and x1=2
x0=3 this expression means that the zero term is 3 and x1=2 means that the first term is 2 correct?
If we need to find 3rd term isn’t it enough to know 2nd term ?
When you solve for x2, where from did you get 3 ?

1. It means that you can apply the given formula to calculate any term starting from x2;
2. Yes.
3. To get x3 you need x2 and x1 because the formula links nth term to TWO proceeding terms, (n-1)st and (n-2)nd

12. Sequences


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The timer says 88% of people got it right but I bet I can count on my fingers the number who did it in two minutes.

Posted from my mobile device
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ScottTargetTestPrep
Walkabout
In a certain sequence, the term \(x_n\) is given by the formula \(x_n=2*x_{n-1}-\frac{1}{2}*x_{n-2}\) for all \(n\geq{2}\). If \(x_0=3\) and \(x_1=2\), what is the value of \(x_3\)?

(A) 2.5
(B) 3.125
(C) 4
(D) 5
(E) 6.75

(Note that the * symbol in both the question stem and in this solution indicates multiplication.)

We are given that X(n) = 2 * X(n-1) – ½ * X(n-2), for all n=>2. This is called a recursive formula, which means that we need to know prior terms before we can compute the subsequent terms. For example, if we want to know X(2), we must know both X(1) and X(0), because X(2) is equal to 2 * X(1) – ½ * X(0).

We are given X(1) = 2 and X(0) = 3. So, when n is 2, X(2) would be calculated as follows:

X(2) = 2 * X(1) – ½ * X(0)

X(2) = 2 * 2 – ½ * 3

X(2) = 4 – 1.5

X(2) = 2.5

Now we are ready to determine the value of X(3). In this case, n = 3, X(1) is 2, and X(2) is 2.5. We plug these values into the recursive formula given in the question stem:

X(3) = 2 * X(2) – ½ * X(1)

X(3) = 2 * 2.5 – ½ * 2

X(3) = 5 – 1

X(3) = 4

Answer is C.

ScottTargetTestPrep
I approached this problem by brute force if you will in that I started by plugging in 3 to X (so x sub 3) to see what I was missing. Is this a bad approach to use when solving other recursive formulas? Thank you :)
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OG Question Code: PS16100
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OG Question Code: PS16100


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Thank you! Added to the first post.
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Hello from the GMAT Club BumpBot!

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