The sum of three consecutive terms of an A.P. is 15 and sum of their squares is 83. Find the terms.

Answer

To solve the problem, we need to find three consecutive terms of an arithmetic progression (A.P.) given that their sum is 15 and the sum of their squares is 83.

### Step 1: Define the terms
Let the three consecutive terms of the A.P. be:
- \( a - d \) (first term)
- \( a \) (second term)
- \( a + d \) (third term)

### Step 2: Set up the equations
From the problem, we have two conditions:
1. The sum of the terms:
\[ (a - d) + a + (a + d) = 15 \]
Simplifying this gives:
\[ 3a = 15 \implies a = 5 \]

2. The sum of the squares of the terms:
\[ (a - d)^2 + a^2 + (a + d)^2 = 83 \]
Expanding this:
\[ (a^2 - 2ad + d^2) + a^2 + (a^2 + 2ad + d^2) = 83 \]
This simplifies to:
\[ 3a^2 + 2d^2 = 83 \]

### Step 3: Substitute \( a \) into the second equation
Now substitute \( a = 5 \) into the equation:
\[ 3(5^2) + 2d^2 = 83 \]
Calculating \( 5^2 \):
\[ 3(25) + 2d^2 = 83 \implies 75 + 2d^2 = 83 \]
Subtracting 75 from both sides:
\[ 2d^2 = 8 \implies d^2 = 4 \implies d = 2 \text{ or } d = -2 \]

### Step 4: Find the terms
Now we can find the three terms:
1. If \( d = 2 \):
- The terms are \( 5 - 2 = 3 \), \( 5 \), and \( 5 + 2 = 7 \).
- So, the terms are \( 3, 5, 7 \).

2. If \( d = -2 \):
- The terms are \( 5 - (-2) = 7 \), \( 5 \), and \( 5 + (-2) = 3 \).
- So, the terms are still \( 3, 5, 7 \).

### Final Answer
The three consecutive terms of the A.P. are \( 3, 5, 7 \).