Question

1. nigel wrote the first six terms of two number patterns below.

pattern a: 1, 2, 3, 4, 5, 6, …………
pattern b: 5, 10, 15, 20, 25, 30, ……
which of the following statements is true about pattern a and pattern b?
a. the 5th, 8th, and 10th term in pattern a will be 5 times the 5th, 8th, and 10th term in pattern b.
b. the 5th, 8th, and 10th term in pattern a will be 5 more than the 5th, 8th, and 10th term in pattern b.
c. the 5th, 8th, and 10th term in pattern b will be 5 times the 5th, 8th, and 10th term in pattern a.
d. the 5th, 8th, and 10th term in pattern b will be 5 more than the 5th, 8th, and 10th term in pattern a.

Explanation:

Step1: Analyze Pattern A

Pattern A: \(1, 2, 3, 4, 5, 6, \dots\)
The \(n\)-th term of Pattern A is \(a_n = n\) (since it's a sequence of natural numbers starting from 1).
So, the 5th term: \(a_5 = 5\)
The 8th term: \(a_8 = 8\)
The 10th term: \(a_{10} = 10\)

Step2: Analyze Pattern B

Pattern B: \(5, 10, 15, 20, 25, 30, \dots\)
The \(n\)-th term of Pattern B is \(b_n = 5n\) (since each term is 5 times its position \(n\)).
So, the 5th term: \(b_5 = 5\times5 = 25\)
The 8th term: \(b_8 = 5\times8 = 40\)
The 10th term: \(b_{10} = 5\times10 = 50\)

Step3: Compare terms of Pattern B with Pattern A

For the 5th term: \(25 = 5\times5\) (5 times the 5th term of A)
For the 8th term: \(40 = 5\times8\) (5 times the 8th term of A)
For the 10th term: \(50 = 5\times10\) (5 times the 10th term of A)
So, the \(n\)-th term of Pattern B is 5 times the \(n\)-th term of Pattern A. Thus, the 5th, 8th, and 10th terms in Pattern B are 5 times those in Pattern A.

Answer:

C. The \(5^{\text{th}}\), \(8^{\text{th}}\), and \(10^{\text{th}}\) term in Pattern B will be 5 times the \(5^{\text{th}}\), \(8^{\text{th}}\), and \(10^{\text{th}}\) term in Pattern A.