Find the sum of an A.P. of 17 terms, whose 3rd term is 8 and 8th term is 28.

Answer

To find the sum of an arithmetic progression (A.P.) with 17 terms, where the 3rd term is 8 and the 8th term is 28, we can follow these steps:

### Step 1: Define the terms of the A.P.
The nth term of an A.P. can be expressed as:
\[ T_n = a + (n-1)d \]
where:
- \( T_n \) is the nth term,
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the term number.

### Step 2: Set up equations using the given terms
From the problem, we know:
- The 3rd term \( T_3 = 8 \):
\[ T_3 = a + (3-1)d = a + 2d = 8 \quad \text{(Equation 1)} \]

- The 8th term \( T_8 = 28 \):
\[ T_8 = a + (8-1)d = a + 7d = 28 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations
Now we have two equations:
1. \( a + 2d = 8 \)
2. \( a + 7d = 28 \)

We can subtract Equation 1 from Equation 2 to eliminate \( a \):
\[ (a + 7d) - (a + 2d) = 28 - 8 \]
This simplifies to:
\[ 5d = 20 \]
Thus, we find:
\[ d = \frac{20}{5} = 4 \]

### Step 4: Substitute \( d \) back to find \( a \)
Now substitute \( d = 4 \) back into Equation 1:
\[ a + 2(4) = 8 \]
\[ a + 8 = 8 \]
\[ a = 8 - 8 = 0 \]

### Step 5: Calculate the sum of the first 17 terms
The formula for the sum \( S_n \) of the first \( n \) terms of an A.P. is:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
For our case:
- \( n = 17 \)
- \( a = 0 \)
- \( d = 4 \)

Substituting these values into the formula:
\[ S_{17} = \frac{17}{2} \times (2(0) + (17-1)(4)) \]
\[ S_{17} = \frac{17}{2} \times (0 + 16 \times 4) \]
\[ S_{17} = \frac{17}{2} \times 64 \]
\[ S_{17} = \frac{17 \times 64}{2} \]
\[ S_{17} = \frac{1088}{2} = 544 \]

### Final Answer
The sum of the 17 terms of the A.P. is **544**.

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