The successive terms of an A.P. are `a_(1), a_(2), a_(3)`,…..
If `a_(6)+a_(9)+a_(12)+a_(15)=20` then `sum_(r=1)^(20)a_(r)`=

Options

A. 75
B. 100
C. 120
D. 150

Answer

To solve the problem step by step, let's denote the first term of the arithmetic progression (A.P.) as \( a_1 = a \) and the common difference as \( d \).

### Step 1: Write the terms of the A.P.
The terms of the A.P. can be expressed as:
- \( a_6 = a + 5d \)
- \( a_9 = a + 8d \)
- \( a_{12} = a + 11d \)
- \( a_{15} = a + 14d \)

### Step 2: Set up the equation based on the given condition
According to the problem, we have:
\[ a_6 + a_9 + a_{12} + a_{15} = 20 \]
Substituting the expressions for \( a_6, a_9, a_{12}, \) and \( a_{15} \):
\[ (a + 5d) + (a + 8d) + (a + 11d) + (a + 14d) = 20 \]

### Step 3: Combine like terms
Combining the terms gives:
\[ 4a + (5d + 8d + 11d + 14d) = 20 \]
Calculating the sum of the coefficients of \( d \):
\[ 5 + 8 + 11 + 14 = 38 \]
Thus, we have:
\[ 4a + 38d = 20 \]

### Step 4: Simplify the equation
Dividing the entire equation by 2:
\[ 2a + 19d = 10 \quad \text{(Equation 1)} \]

### Step 5: Find the sum of the first 20 terms of the A.P.
The sum \( S_n \) of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
For \( n = 20 \):
\[ S_{20} = \frac{20}{2} \times (2a + 19d) = 10 \times (2a + 19d) \]

### Step 6: Substitute the value from Equation 1
From Equation 1, we know that \( 2a + 19d = 10 \). Substituting this into the sum formula:
\[ S_{20} = 10 \times 10 = 100 \]

### Final Answer
Thus, the sum of the first 20 terms of the A.P. is:
\[ \boxed{100} \]