The sum of the fourth and twelfth term of an arithmetic progression is 20. What is the sum of the first 15 terms of the arithmetic progression?

Answer

To solve the problem step by step, we will use the formulas for the nth term and the sum of the first n terms of an arithmetic progression (AP).

### Step 1: Understand the Problem
We are given that the sum of the fourth and twelfth terms of an arithmetic progression is 20. We need to find the sum of the first 15 terms of this AP.

### Step 2: Use the Formula for the nth Term of an AP
The nth term of an arithmetic progression can be expressed as:
\[ A_n = A + (n - 1)D \]
where:
- \( A \) is the first term,
- \( D \) is the common difference,
- \( n \) is the term number.

For the fourth term (\( A_4 \)):
\[ A_4 = A + (4 - 1)D = A + 3D \]

For the twelfth term (\( A_{12} \)):
\[ A_{12} = A + (12 - 1)D = A + 11D \]

### Step 3: Set Up the Equation
According to the problem, the sum of the fourth and twelfth terms is 20:
\[ A_4 + A_{12} = 20 \]
Substituting the expressions for \( A_4 \) and \( A_{12} \):
\[ (A + 3D) + (A + 11D) = 20 \]

### Step 4: Simplify the Equation
Combine like terms:
\[ 2A + 14D = 20 \]
This is our Equation 1.

### Step 5: Use the Formula for the Sum of the First n Terms of an AP
The sum of the first \( n \) terms of an arithmetic progression is given by:
\[ S_n = \frac{n}{2} \times (2A + (n - 1)D) \]
For the sum of the first 15 terms (\( S_{15} \)):
\[ S_{15} = \frac{15}{2} \times (2A + (15 - 1)D) = \frac{15}{2} \times (2A + 14D) \]

### Step 6: Substitute from Equation 1
From Equation 1, we know that \( 2A + 14D = 20 \). Substitute this into the equation for \( S_{15} \):
\[ S_{15} = \frac{15}{2} \times 20 \]

### Step 7: Calculate \( S_{15} \)
Now calculate:
\[ S_{15} = \frac{15 \times 20}{2} = \frac{300}{2} = 150 \]

### Final Answer
The sum of the first 15 terms of the arithmetic progression is:
\[ S_{15} = 150 \]

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