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 $8,12,2929,42,\cdots$

To solve for the 6th term ($p$) and 7th term ($q$) of the sequence $8, 12, 29, 42, \cdots$, 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.

\[
\text{First difference} = (12 - 8), (29 - 12), (42 - 29), \cdots
\]

\[
\Rightarrow 4, 17, 13, \cdots
\]

### Step 2: Compute the second differences.
The **second difference** is the difference between consecutive values of the first differences.

\[
\text{Second difference} = (17 - 4), (13 - 17), \cdots
\]

\[
\Rightarrow 13, -4, \cdots
\]

### 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 $-4$.

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

### Step 4: Extend the first differences.
To find the next first differences, we add the second differences to the existing first differences:

- After $13$, add $-4$:
  \[
  13 + (-4) = 9
  \]
- After $9$, add $-4$:
  \[
  9 + (-4) = 5
  \]

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

### 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:
  \[
  42 + 9 = 51
  \]
- The 6th term ($p$) is:
  \[
  51 + 5 = 56
  \]

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

Now, for the 7th term ($q$):

- The next first difference is $5 + (-4) = 1$.
- The 7th term ($q$) is:
  \[
  56 + 1 = 57
  \]

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

### Final Answer:
- $p = 56$ (the 6th term)
- $q = 57$ (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.

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, 29, 42, ...

Learn how to find the 6th and 7th terms of the sequence 8, 12, 29, 42, ... by constructing tables and using second-order differences. This step-by-step solution covers key concepts like sequences, first and second differences, and algebraic methods. Ideal for students in Grades 9-10.

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