# Number Sequence Review Tutorial and Practice Questions

- Posted by Brian Stocker MA
- Date Published June 20, 2014
- Date modified May 21, 2020
- Comments 0 comment

### Sequences

A *sequence*, in mathematics, is a string of objects, like *numbers*, that follow a pattern. The individual elements in a sequence are called terms.

Number sequence questions as for a particular term in the sequence, which can be calculated or reasons from the pattern.

Number sequences are College Level Math and appear on standardized tests like the Accuplacer and the COMPASS.

### Review and Tutorial

A sequence of numbers is a set of numbers, but here they are in order. For example, we can represent the set of natural numbers N as a sequence 1, 2, 3,… A sequence can be finite or infinite. In our case of the sequence of the natural numbers, we have an infinite sequence.

If we have a sequence of numbers a_{1}, a_{2}, a_{3}, … we denote that sequence by {a_{n}}. We can write, for example, the sequence of natural numbers like this:

a_{n} = a_{n-1} + 1 or a_{n+1} = a_{n+1}

From this formula, we can see that each number is greater than the previous number by one, which is true for the sequence of the natural numbers.

The first term (member) of the sequence is denoted by a_{0}. So, if we know the first term of the sequence and we know the formula that describes the sequence, we can find any term of that sequence. Even if we know some other member of the sequence, we can find other members.

**Let’s solve 2 examples for both cases:**

1) If a_{0} = 2 and an = an-1 – 2, find the 4th member of the sequence {a_{n}}.

Let’s find 2nd and 3rd member, which we will use to find the 4th.

a_{1} = a_{0} – 2 = 2 – 2 = 0

a_{2} = a_{1} – 2 = 0 – 2 = -2

a_{3} = a_{2} – 2 = -2 – 2 = -4

So, our 4th member is number -4.

2) If a_{2} = 4 and a_{n }= 2a_{n-1}, find the 1st member of the sequence {a_{n}}.

a_{2} = 2a_{1} → 4 = 2a_{1} → a_{1} = 2

a_{1} = 2 a_{0} → 2 = 2a_{0} → a_{0} = 1

So, our first member is 1.

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**Practice Questions**

**1. If a_{0 }= 3 and a_{n }= – a_{n-1 }+ 3, find a_{3} of the sequence {a_{n}}.**

A. 0

B. 1

C. 2

D. 3

**2. If terms of the sequence {a_{n}} are represented by a_{n }= a_{n-1}/n and a_{1 }= 1, find a_{4}.**

A. 1/2

B. 1/4

C. 1/16

D. 1/24

**3. If a_{0 }= 1/2 and a_{n }= 2a_{n-1}^{2} , find a_{2} of the sequence {a_{n}}.**

A. 1/2

B. 1/4

C. 1/16

D. 1/24

**4. If members of the sequence {a_{n}} are represented by a_{n }= (-1)^{n}a_{n-1} and if a_{2 }= 2, find a_{0}.**

A. 2

B. 1

C. 0

D. -2

**5. If the sequence {a_{n}} is defined by a_{n+1 }= 1- a_{n} and a_{2 }= 6, find a_{4}.**

A. 2

B. 1

C. 6

D. -1

**6. If members of the sequence {a_{n}} are represented by a_{n+1 }= – a_{n-1} and a_{2 }= 3 and, find a_{3 }+ a_{4}.**

A. 2

B. 3

C. 0

D. -2

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### Answer Key

**1. A. 0**

*a _{0 }= 3*

*a _{n }= – a_{n-1 }+ 3*

*a _{1 }= – a_{0 }+ 3 = -3 + 3 = 0*

*a _{2 }= – a_{1 }+ 3 = 0 + 3 = 3*

*a _{3 }= – a_{2 }+ 3 = -3 + 3 = 0*

**2. D. 1/24**

*a _{n }= a_{n-1}/n*

*a _{1 }= 1*

*a _{2 }= a_{1}/2 = 1/2*

*a _{3 }= a_{2}/3 = (1/2)/3 = 1/6*

*a _{4 }= a_{3}/4 = (1/6)/4 = 1/24*

**3. A. 1/2**

*a _{0 }= 1/2*

*a _{n }= 2a_{n-1}^{2}*

*a _{1 }= 2a_{0}^{2 }= 2·(1/2)^{2 }= 2·(1/4) = 1/2*

*a _{2 }= 2a_{1}^{2 }= 2·(1/2)^{2 }= 2·(1/4) = 1/2*

**4. D -2**

*a _{n}= (-1)^{n}a_{n-1}*

*a _{2}=2*

*2=a _{2}= (-1)^{2}a_{1}= a_{1} ? a_{1}=2*

*a _{1}= (-1)^{1}a_{0}*

*2=- a _{0}*

*a _{0}=-2*

**5. C. 6**

*a _{n+1 }= 1 – a_{n}*

*a _{2 }= 6*

*a _{3 }= 1 – a_{2 }= 1 – 6 = -5*

*a _{4 }= 1 – a_{3 }= 1 – (-5) = 1 + 5 = 6*

**6. C. 0**

*a _{n+1 }= – a_{n-1}*

*a _{2 }= 3*

*a _{3 }= – a_{2 }= -3*

*a _{4 }= – a_{3 }= -(-3) = 3*

*a _{3 }+ a_{4 }= -3 + 3 = 0*

**Written by:**Brian Stocker MA, Complete Test Preparation Inc.