The sum of three consecutive terms of an A.P. is 21 and the sum of their squares is 165. Find these terms.

Answer

To solve the problem, we need to find three consecutive terms of an Arithmetic Progression (A.P.) given two conditions: their sum is 21, and the sum of their squares is 165.

### Step-by-Step Solution

**Step 1: Define the Terms**
Let the three consecutive terms of the A.P. be:
- First term: \( a - d \)
- Second term: \( a \)
- Third term: \( a + d \)

**Step 2: Set Up the Equation for the Sum**
According to the problem, the sum of these terms is:
\[ (a - d) + a + (a + d) = 21 \]
This simplifies to:
\[ 3a = 21 \]

**Step 3: Solve for \( a \)**
Dividing both sides by 3 gives:
\[ a = \frac{21}{3} = 7 \]

**Step 4: Set Up the Equation for the Sum of Squares**
Next, we use the second condition regarding the sum of their squares:
\[ (a - d)^2 + a^2 + (a + d)^2 = 165 \]
Expanding this, we have:
\[ (a^2 - 2ad + d^2) + a^2 + (a^2 + 2ad + d^2) = 165 \]
Combining like terms results in:
\[ 3a^2 + 2d^2 = 165 \]

**Step 5: Substitute the Value of \( a \)**
Now we substitute \( a = 7 \) into the equation:
\[ 3(7^2) + 2d^2 = 165 \]
Calculating \( 7^2 \):
\[ 3(49) + 2d^2 = 165 \]
This simplifies to:
\[ 147 + 2d^2 = 165 \]

**Step 6: Solve for \( d^2 \)**
Subtracting 147 from both sides gives:
\[ 2d^2 = 165 - 147 \]
\[ 2d^2 = 18 \]
Dividing by 2:
\[ d^2 = 9 \]

**Step 7: Solve for \( d \)**
Taking the square root gives:
\[ d = \pm 3 \]

**Step 8: Find the Terms**
Now we can find the three terms:
1. If \( d = 3 \):
- First term: \( a - d = 7 - 3 = 4 \)
- Second term: \( a = 7 \)
- Third term: \( a + d = 7 + 3 = 10 \)

2. If \( d = -3 \):
- First term: \( a - d = 7 - (-3) = 7 + 3 = 10 \)
- Second term: \( a = 7 \)
- Third term: \( a + d = 7 + (-3) = 7 - 3 = 4 \)

Thus, the three terms of the A.P. are \( 4, 7, 10 \) or \( 10, 7, 4 \).

### Final Answer
The three terms of the A.P. are \( 4, 7, 10 \) or \( 10, 7, 4 \).