Math Problem Statement

by constructing a different table and using the second order differences as constant, find the sixth term(p) and the seventh term(q) of the sequence $8,12,2929,42,\cdots$

Solution

To solve for the 6th term (pp) and 7th term (qq) of the sequence 8,12,29,42,8, 12, 29, 42, \cdots, using the method of second-order differences, we'll follow these steps:

Step 1: Compute the first differences.

The first difference is the difference between consecutive terms in the sequence.

First difference=(128),(2912),(4229),\text{First difference} = (12 - 8), (29 - 12), (42 - 29), \cdots

4,17,13,\Rightarrow 4, 17, 13, \cdots

Step 2: Compute the second differences.

The second difference is the difference between consecutive values of the first differences.

Second difference=(174),(1317),\text{Second difference} = (17 - 4), (13 - 17), \cdots

13,4,\Rightarrow 13, -4, \cdots

Step 3: Assume the second differences are constant.

We are told that the second differences remain constant, so we assume the second difference continues to be 4-4.

Thus, the second differences are: 13,4,4,13, -4, -4, \cdots.

Step 4: Extend the first differences.

To find the next first differences, we add the second differences to the existing first differences:

  • After 1313, add 4-4: 13+(4)=913 + (-4) = 9
  • After 99, add 4-4: 9+(4)=59 + (-4) = 5

Thus, the first differences become: 4,17,13,9,54, 17, 13, 9, 5.

Step 5: Extend the sequence.

To find the 6th and 7th terms, we add the first differences to the current sequence values:

  • The 5th term (after 42) is: 42+9=5142 + 9 = 51
  • The 6th term (pp) is: 51+5=5651 + 5 = 56

Thus, the 6th term is p=56p = 56.

Now, for the 7th term (qq):

  • The next first difference is 5+(4)=15 + (-4) = 1.
  • The 7th term (qq) is: 56+1=5756 + 1 = 57

Thus, the 7th term is q=57q = 57.

Final Answer:

  • p=56p = 56 (the 6th term)
  • q=57q = 57 (the 7th term)

Would you like further details or have any questions?

Here are 5 related questions:

  1. How can second-order differences be applied to quadratic sequences?
  2. What if the second-order differences were not constant? How would that affect the solution?
  3. Can you explain the relationship between first, second, and third-order differences?
  4. How can you use this method for sequences with fractional terms?
  5. What are some real-world applications of difference methods?

Tip: Second-order differences suggest that the underlying rule of the sequence may be quadratic in nature.

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Math Problem Analysis

Mathematical Concepts

Sequences
Second-order Differences
Algebra

Formulas

First differences: T(n) = T(n+1) - T(n)
Second differences: Δ² = Δ(T(n+1)) - Δ(T(n))

Theorems

Difference Method for Polynomial Sequences

Suitable Grade Level

Grades 9-10