The sum of three consecutive terms of an A.P. is `9` and the sum of their squares is `35`. Find the terms.

Answer

To solve the problem, we need to find three consecutive terms of an arithmetic progression (A.P.) given two conditions: the sum of the terms is 9, and the sum of their squares is 35.

### Step-by-Step Solution:

1. **Define the Terms**:
Let the three consecutive terms of the A.P. be:
\[ A - D, A, A + D \]
where \( A \) is the middle term and \( D \) is the common difference.

2. **Set Up the First Equation**:
According to the problem, the sum of these terms is 9:
\[ (A - D) + A + (A + D) = 9 \]
Simplifying this, we get:
\[ 3A = 9 \]
Therefore, dividing both sides by 3:
\[ A = 3 \]

3. **Set Up the Second Equation**:
Next, we consider the sum of the squares of these terms:
\[ (A - D)^2 + A^2 + (A + D)^2 = 35 \]
Expanding this:
\[ (A^2 - 2AD + D^2) + A^2 + (A^2 + 2AD + D^2) = 35 \]
Combining like terms:
\[ 3A^2 + 2D^2 = 35 \]

4. **Substitute the Value of A**:
Now substitute \( A = 3 \) into the equation:
\[ 3(3^2) + 2D^2 = 35 \]
Calculating \( 3^2 \):
\[ 3 \times 9 + 2D^2 = 35 \]
This simplifies to:
\[ 27 + 2D^2 = 35 \]

5. **Solve for D**:
Rearranging gives:
\[ 2D^2 = 35 - 27 \]
Simplifying further:
\[ 2D^2 = 8 \]
Dividing by 2:
\[ D^2 = 4 \]
Taking the square root of both sides:
\[ D = \pm 2 \]

6. **Find the Terms**:
Now we can find the terms of the A.P. for both cases of \( D \):
- If \( D = 2 \):
\[ \text{Terms: } A - D = 3 - 2 = 1, \quad A = 3, \quad A + D = 3 + 2 = 5 \]
So the terms are \( 1, 3, 5 \).

- If \( D = -2 \):
\[ \text{Terms: } A - D = 3 - (-2) = 5, \quad A = 3, \quad A + D = 3 + (-2) = 1 \]
So the terms are \( 5, 3, 1 \).

### Final Answer:
The three consecutive terms of the A.P. are \( 1, 3, 5 \) or \( 5, 3, 1 \).