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 follow the reasoning laid out in the video transcript.

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

### Step 2: Use the General Term Formula
The nth term of an arithmetic progression can be expressed using the formula:
\[ a_n = a + (n - 1)d \]
where:
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the term number.

### Step 3: Write the Expressions for the 4th and 12th Terms
Using the formula for the 4th term (\( n = 4 \)):
\[ a_4 = a + (4 - 1)d = a + 3d \]

Using the formula for the 12th term (\( n = 12 \)):
\[ a_{12} = a + (12 - 1)d = a + 11d \]

### Step 4: Set Up the Equation
According to the problem, the sum of the 4th and 12th terms is 20:
\[ a_4 + a_{12} = 20 \]
Substituting the expressions we found:
\[ (a + 3d) + (a + 11d) = 20 \]
This simplifies to:
\[ 2a + 14d = 20 \]

### Step 5: Simplify the Equation
We can simplify the equation:
\[ 2a + 14d = 20 \]
Dividing the entire equation by 2:
\[ a + 7d = 10 \]
This is our Equation (1).

### Step 6: Use the Sum Formula for the First 15 Terms
The sum of the first \( n \) terms of an arithmetic progression is given by the formula:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
For the first 15 terms (\( n = 15 \)):
\[ S_{15} = \frac{15}{2} \times (2a + (15 - 1)d) \]
This simplifies to:
\[ S_{15} = \frac{15}{2} \times (2a + 14d) \]

### Step 7: Substitute from Equation (1)
From Equation (1), we know:
\[ 2a + 14d = 20 \]
Substituting this into the sum formula:
\[ S_{15} = \frac{15}{2} \times 20 \]

### Step 8: Calculate the Sum
Now, we calculate:
\[ S_{15} = \frac{15 \times 20}{2} = \frac{300}{2} = 150 \]

### Conclusion
The sum of the first 15 terms of the arithmetic progression is:
\[ \boxed{150} \]

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