Find the Pattern Rule: Analyzing Sequence 3,15,27,39,41 Using Variable n

Question

Given the series whose difference between two jumped numbers is constant:

3,15,27,39,41 3,15,27,39,41

Describe the property using the variable n n

Video Solution

Solution Steps

00:00 Find the sequence formula
00:05 This is the first term according to the given data
00:10 Let's identify the difference between terms (D) according to the given data
00:22 We'll use the formula to describe the sequence
00:32 We'll substitute appropriate values and solve to find the sequence formula
00:41 And this is the solution to the question

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Recognize the pattern in the series.
  • Step 2: Identify the first term and the common difference.
  • Step 3: Derive the formula for the sequence.

Now, let's work through each step:

Step 1: The given series is 3,15,27,39,413, 15, 27, 39, 41. Observing the first few terms, 3,15,27,393, 15, 27, 39, we note each subsequent number increases by 1212, forming an arithmetic sequence. The number 4141 does not follow this arithmetic sequence pattern.

Step 2: The first term a1a_1 is 33, and the common difference dd is 1212, as derived from verifying the difference between each two successive terms.

Step 3: We use the arithmetic sequence formula:
a(n)=a1+(n1)×d\displaystyle a(n) = a_1 + (n-1) \times d

Substitute the known values:
a(n)=3+(n1)×12\displaystyle a(n) = 3 + (n-1) \times 12

This formula describes the arithmetic sequence for the original numbers 3,15,27,393, 15, 27, 39 but not for 4141.

Therefore, the solution to the problem is:

a(n)=3+(n1)×12 a(n) = 3 + (n-1) \times 12 .

Answer

a(n)=3+(n1)×12 a(n)=3+(n-1)\times12