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Number - Number sequences (prediction, description)

Grade 10IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A sequence is an ordered list of numbers where each number is called a term. The position of a term is denoted by nn (where n=1,2,3,…n=1, 2, 3, \dots).

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An Arithmetic Sequence (Linear) has a constant difference between consecutive terms, called the common difference (dd). It follows a rule of the form un=dn+cu_n = dn + c.

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A Geometric Sequence has a constant multiplier between consecutive terms, called the common ratio (rr). Each term is found by multiplying the previous term by rr.

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A Quadratic Sequence is one where the first differences are not constant, but the second differences are constant. The nthn^{th} term formula is of the form un=an2+bn+cu_n = an^2 + bn + c.

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Special Sequences include Square numbers (1,4,9,16…1, 4, 9, 16 \dots), Cube numbers (1,8,27,64…1, 8, 27, 64 \dots), and the Fibonacci sequence (1,1,2,3,5,8…1, 1, 2, 3, 5, 8 \dots) where each term is the sum of the two preceding ones.

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The nthn^{th} term, denoted as unu_n or TnT_n, allows us to calculate the value of a term at any position without listing all previous terms.

📐Formulae

un=a+(n−1)du_n = a + (n-1)d (General term of an Arithmetic Sequence)

d=un−un−1d = u_n - u_{n-1} (Common difference)

un=arn−1u_n = ar^{n-1} (General term of a Geometric Sequence)

r=unun−1r = \frac{u_n}{u_{n-1}} (Common ratio)

un=an2+bn+cu_n = an^2 + bn + c (General term of a Quadratic Sequence, where 2a2a is the constant second difference)

💡Examples

Problem 1:

Find the nthn^{th} term formula for the arithmetic sequence: 7,11,15,19,…7, 11, 15, 19, \dots

Solution:

un=4n+3u_n = 4n + 3

Explanation:

First, find the common difference dd: 11−7=411 - 7 = 4. The formula follows un=dn+cu_n = dn + c, so un=4n+cu_n = 4n + c. To find cc, use the first term (n=1n=1): 4(1)+c=74(1) + c = 7, which gives c=3c = 3. Therefore, un=4n+3u_n = 4n + 3.

Problem 2:

Determine the 10th10^{th} term of the geometric sequence: 3,6,12,24,…3, 6, 12, 24, \dots

Solution:

u10=1536u_{10} = 1536

Explanation:

Identify the first term a=3a = 3 and the common ratio r=63=2r = \frac{6}{3} = 2. Use the formula un=arn−1u_n = ar^{n-1}. For the 10th10^{th} term: u10=3×210−1=3×29=3×512=1536u_{10} = 3 \times 2^{10-1} = 3 \times 2^9 = 3 \times 512 = 1536.

Problem 3:

Find the nthn^{th} term of the sequence: 4,9,16,25,…4, 9, 16, 25, \dots

Solution:

un=(n+1)2u_n = (n+1)^2

Explanation:

Observe that these are square numbers. The standard square numbers are 12,22,32,…1^2, 2^2, 3^2, \dots. Comparing 1,4,9,161, 4, 9, 16 with 4,9,16,254, 9, 16, 25, we see the sequence is shifted. When n=1,u1=(1+1)2=4n=1, u_1 = (1+1)^2 = 4. When n=2,u2=(2+1)2=9n=2, u_2 = (2+1)^2 = 9. Thus, un=(n+1)2u_n = (n+1)^2.

Problem 4:

Find the next two terms in the sequence: 1,3,6,10,…1, 3, 6, 10, \dots

Solution:

15,2115, 21

Explanation:

This is the sequence of Triangle Numbers. The difference between terms increases by 1 each time: 3−1=23-1=2, 6−3=36-3=3, 10−6=410-6=4. The next difference will be 55 (10+5=1510+5=15) and the one after will be 66 (15+6=2115+6=21).