The sum of the first 30 terms of an A.P. is 1920. if the fourth term is 18, find its `11^(th)` term.

Answer

To solve the problem step by step, we will use the formulas for the sum of an arithmetic progression (A.P.) and the nth term of an A.P.

### Step 1: Understand the given information
We are given:
- The sum of the first 30 terms of an A.P. (S30) = 1920
- The fourth term (a4) = 18

We need to find the 11th term (a11).

### Step 2: Use the formula for the sum of the first n terms of an A.P.
The formula for the sum of the first n terms of an A.P. is:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
For our case, n = 30, so we have:
\[ S_{30} = \frac{30}{2} \times (2a + (30 - 1)d) \]
Substituting the known value:
\[ 1920 = 15 \times (2a + 29d) \]

### Step 3: Simplify the equation
Dividing both sides by 15:
\[ 128 = 2a + 29d \] (Equation 1)

### Step 4: Use the formula for the nth term of an A.P.
The formula for the nth term of an A.P. is:
\[ a_n = a + (n - 1)d \]
For the fourth term:
\[ a_4 = a + 3d \]
We know that a4 = 18, so:
\[ 18 = a + 3d \] (Equation 2)

### Step 5: Solve the system of equations
Now we have two equations:
1. \( 2a + 29d = 128 \) (Equation 1)
2. \( a + 3d = 18 \) (Equation 2)

From Equation 2, we can express a in terms of d:
\[ a = 18 - 3d \]

### Step 6: Substitute a in Equation 1
Substituting \( a \) in Equation 1:
\[ 2(18 - 3d) + 29d = 128 \]
Expanding this:
\[ 36 - 6d + 29d = 128 \]
Combining like terms:
\[ 36 + 23d = 128 \]
Subtracting 36 from both sides:
\[ 23d = 92 \]
Dividing by 23:
\[ d = 4 \]

### Step 7: Find the value of a
Now substitute \( d = 4 \) back into Equation 2 to find \( a \):
\[ a + 3(4) = 18 \]
\[ a + 12 = 18 \]
Subtracting 12 from both sides:
\[ a = 6 \]

### Step 8: Find the 11th term (a11)
Using the formula for the nth term:
\[ a_{11} = a + (11 - 1)d \]
Substituting the values of \( a \) and \( d \):
\[ a_{11} = 6 + 10 \times 4 \]
\[ a_{11} = 6 + 40 \]
\[ a_{11} = 46 \]

### Final Answer
The 11th term of the A.P. is **46**.

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