To solve the problem, we need to find three numbers in an arithmetic progression (AP) whose sum is 3 and whose product is -35.
### Step-by-Step Solution:
1. **Define the Numbers**:
Let the three numbers in AP be:
\( A - D, A, A + D \)
where \( A \) is the middle term and \( D \) is the common difference.
**Hint**: Remember that in an AP, the middle term is the average of the other two terms.
2. **Set Up the Sum Equation**:
According to the problem, the sum of these three numbers is 3:
\[
(A - D) + A + (A + D) = 3
\]
Simplifying this, we get:
\[
3A = 3
\]
**Hint**: Combine like terms carefully to isolate \( A \).
3. **Solve for A**:
Dividing both sides by 3 gives:
\[
A = 1
\]
**Hint**: Make sure to check your calculations when solving for a variable.
4. **Set Up the Product Equation**:
The product of the three numbers is given as -35:
\[
(A - D) \cdot A \cdot (A + D) = -35
\]
Substituting \( A = 1 \) into the equation:
\[
(1 - D) \cdot 1 \cdot (1 + D) = -35
\]
This simplifies to:
\[
(1 - D)(1 + D) = -35
\]
**Hint**: Remember that \( (a - b)(a + b) = a^2 - b^2 \).
5. **Simplify the Product**:
Expanding the left side gives:
\[
1^2 - D^2 = -35
\]
Thus, we have:
\[
1 - D^2 = -35
\]
**Hint**: Rearranging equations can help isolate the variable you need.
6. **Rearrange to Solve for D**:
Adding \( D^2 \) to both sides and adding 35 to both sides gives:
\[
D^2 = 36
\]
**Hint**: When solving for squares, consider both the positive and negative roots.
7. **Find D**:
Taking the square root of both sides yields:
\[
D = 6 \quad \text{or} \quad D = -6
\]
**Hint**: Remember that both positive and negative values of \( D \) will give valid sequences.
8. **Calculate the Numbers**:
Now we substitute \( A \) and \( D \) back into our expressions for the three numbers:
- For \( D = 6 \):
\[
A - D = 1 - 6 = -5, \quad A = 1, \quad A + D = 1 + 6 = 7
\]
The numbers are: \( -5, 1, 7 \).
- For \( D = -6 \):
\[
A - D = 1 - (-6) = 1 + 6 = 7, \quad A = 1, \quad A + D = 1 + (-6) = -5
\]
The numbers are: \( 7, 1, -5 \).
**Hint**: Make sure to check that both sets of numbers satisfy the original conditions.
### Final Answer:
The three numbers in AP are \( -5, 1, 7 \) (or \( 7, 1, -5 \) depending on the order).
