Let's denote the three consecutive terms of an arithmetic progression (AP) as $a - d$, $a$, and $a + d$, where $a$ is the middle term and $d$ is the common difference.

Step 1: Form Equations from the Given Information

1. Sum of the Three Terms: $(a - d) + a + (a + d) = 33$ Simplifying: $3a = 33$ $a = 11$

2. Product of the Three Terms: $(a - d) \times a \times (a + d) = 1287$ Substituting $a = 11$: $(11 - d) \times 11 \times (11 + d) = 1287$ This simplifies to: $(11^2 - d^2) \times 11 = 1287$ $121 - d^2 = \frac{1287}{11} = 117$ $d^2 = 121 - 117 = 4$ $d = \pm 2$

Step 2: Determine the Three Numbers

• If $d = 2$, the three terms are: $11 - 2 = 9, \quad 11, \quad 11 + 2 = 13$ So the three numbers are 9, 11, and 13.

• If $d = -2$, the three terms would be: $11 + 2 = 13, \quad 11, \quad 11 - 2 = 9$ However, these are the same numbers just in a different order.

Thus, the three consecutive numbers are 9, 11, and 13.

Step 3: Find the 7th and 14th Terms of the AP

The general formula for the $n$th term of an AP is:

$a_n = a_1 + (n - 1) \times d$

Where:

• $a_1$ is the first term (which is 9),
• $d$ is the common difference (which is 2),
• $n$ is the term number.

1. 7th Term ($a_7$): $a_7 = 9 + (7 - 1) \times 2 = 9 + 6 \times 2 = 9 + 12 = 21$

2. 14th Term ($a_{14}$): $a_{14} = 9 + (14 - 1) \times 2 = 9 + 13 \times 2 = 9 + 26 = 35$

Final Results:

• The three consecutive numbers are 9, 11, and 13.
• The 7th term of the AP is 21.
• The 14th term of the AP is 35.

Would you like more details or have any questions?

Related Questions:

1. What would be the 10th term of this AP?
2. How would the AP change if the common difference were negative?
3. What is the sum of the first 10 terms of this AP?
4. What if the sum of four consecutive terms of an AP was given instead?
5. How does the common difference affect the rate at which the terms increase?

Tip: In an arithmetic progression, the common difference remains constant, which makes it easy to predict future terms based on earlier ones.