The 17th term of AP is 5 more than twice its 8th term. If the 11th term of the AP is 43, find its nth term.

Explanation

To solve the problem step by step, we will use the properties of an Arithmetic Progression (AP).

### Step 1: Understand the given information
We know:
1. The 17th term of the AP is 5 more than twice the 8th term.
2. The 11th term of the AP is 43.

### Step 2: Write the general formula for the nth term of an AP
The nth term of an AP can be expressed as:
\[ T_n = A + (n-1)D \]
where:
- \( A \) is the first term,
- \( D \) is the common difference,
- \( n \) is the term number.

### Step 3: Express the 17th and 8th terms using the formula
Using the formula, we can express the 17th and 8th terms:
- The 17th term \( T_{17} \):
\[ T_{17} = A + (17-1)D = A + 16D \]
- The 8th term \( T_{8} \):
\[ T_{8} = A + (8-1)D = A + 7D \]

### Step 4: Set up the equation from the first condition
According to the problem, the 17th term is 5 more than twice the 8th term:
\[ T_{17} = 2T_{8} + 5 \]
Substituting the expressions we found:
\[ A + 16D = 2(A + 7D) + 5 \]

### Step 5: Simplify the equation
Expanding the right side:
\[ A + 16D = 2A + 14D + 5 \]
Rearranging gives:
\[ A + 16D - 2A - 14D = 5 \]
This simplifies to:
\[ -A + 2D = 5 \]
or
\[ A - 2D = -5 \] (Equation 1)

### Step 6: Express the 11th term using the formula
Now, we can express the 11th term:
\[ T_{11} = A + (11-1)D = A + 10D \]
According to the problem, this equals 43:
\[ A + 10D = 43 \] (Equation 2)

### Step 7: Solve the system of equations
Now we have two equations:
1. \( A - 2D = -5 \)
2. \( A + 10D = 43 \)

We can solve these equations simultaneously. Let's subtract Equation 1 from Equation 2:
\[ (A + 10D) - (A - 2D) = 43 - (-5) \]
This simplifies to:
\[ 12D = 48 \]
Thus, we find:
\[ D = 4 \]

### Step 8: Substitute back to find A
Now substitute \( D = 4 \) back into Equation 1:
\[ A - 2(4) = -5 \]
\[ A - 8 = -5 \]
Thus, we find:
\[ A = 3 \]

### Step 9: Write the general term of the AP
Now that we have \( A \) and \( D \), we can write the nth term of the AP:
\[ T_n = A + (n-1)D \]
Substituting the values of \( A \) and \( D \):
\[ T_n = 3 + (n-1) \cdot 4 \]
This simplifies to:
\[ T_n = 3 + 4n - 4 \]
Thus:
\[ T_n = 4n - 1 \]

### Final Answer
The nth term of the AP is:
\[ T_n = 4n - 1 \]