The first four terms of an AP are a, 9, `3a- b, 3a+b`.
The 2011 th term of an AP is

Options

A. 2015
B. 4025
C. 5030
D. 8045

Explanation

To solve the problem, we need to find the 2011th term of the arithmetic progression (AP) given the first four terms: \( a, 9, 3a - b, 3a + b \).

### Step 1: Set up the equations based on the property of AP
In an AP, the difference between consecutive terms is constant. Therefore, we can set up the following equations based on the first four terms:

1. The difference between the second term and the first term:
\[ 9 - a = (3a - b) - 9 \]

2. The difference between the third term and the second term:
\[ (3a - b) - 9 = (3a + b) - (3a - b) \]

### Step 2: Solve the first equation
From the first equation:
\[ 9 - a = 3a - b - 9 \]
Rearranging gives:
\[ 9 - a + 9 = 3a - b \]
\[ 18 - a = 3a - b \]
This simplifies to:
\[ 18 = 4a - b \quad \text{(Equation 1)} \]

### Step 3: Solve the second equation
From the second equation:
\[ (3a - b) - 9 = (3a + b) - (3a - b) \]
This simplifies to:
\[ 3a - b - 9 = 2b \]
Rearranging gives:
\[ 3a - b - 9 = 2b \]
\[ 3a - 9 = 3b \]
This simplifies to:
\[ 3a - 9 = 3b \quad \text{or} \quad a - 3 = b \quad \text{(Equation 2)} \]

### Step 4: Substitute Equation 2 into Equation 1
Now we substitute \( b = a - 3 \) into Equation 1:
\[ 18 = 4a - (a - 3) \]
This simplifies to:
\[ 18 = 4a - a + 3 \]
\[ 18 = 3a + 3 \]
Subtracting 3 from both sides:
\[ 15 = 3a \]
Dividing by 3:
\[ a = 5 \]

### Step 5: Find the value of b
Now substitute \( a = 5 \) back into Equation 2:
\[ b = 5 - 3 = 2 \]

### Step 6: Identify the first term and common difference
Now we have:
- \( a = 5 \)
- \( b = 2 \)

The first four terms of the AP are:
- First term: \( a = 5 \)
- Second term: \( 9 \)
- Third term: \( 3a - b = 3(5) - 2 = 15 - 2 = 13 \)
- Fourth term: \( 3a + b = 3(5) + 2 = 15 + 2 = 17 \)

The common difference \( d \) can be calculated as:
\[ d = 9 - 5 = 4 \]

### Step 7: Find the 2011th term
The formula for the nth term of an AP is given by:
\[ T_n = a + (n - 1)d \]
Substituting \( n = 2011 \):
\[ T_{2011} = 5 + (2011 - 1) \cdot 4 \]
\[ T_{2011} = 5 + 2010 \cdot 4 \]
\[ T_{2011} = 5 + 8040 \]
\[ T_{2011} = 8045 \]

### Final Answer
The 2011th term of the AP is **8045**.