The 6th term of an AP. is 16 and the 14th term is 32. Determine the 36th term.

Answer

To find the 36th term of an arithmetic progression (AP) where the 6th term is 16 and the 14th term is 32, we can follow these steps:

### Step 1: Write down the given information.
We know:
- The 6th term \( A_6 = 16 \)
- The 14th term \( A_{14} = 32 \)

### Step 2: Use the formula for the nth term of an AP.
The formula for the nth term of an AP is given by:
\[ A_n = A + (n - 1)D \]
where:
- \( A \) is the first term,
- \( D \) is the common difference,
- \( n \) is the term number.

### Step 3: Write equations for the given terms.
Using the formula:
1. For the 6th term:
\[ A_6 = A + (6 - 1)D = A + 5D = 16 \quad \text{(Equation 1)} \]
2. For the 14th term:
\[ A_{14} = A + (14 - 1)D = A + 13D = 32 \quad \text{(Equation 2)} \]

### Step 4: Set up the equations.
From Equation 1:
\[ A + 5D = 16 \]
From Equation 2:
\[ A + 13D = 32 \]

### Step 5: Subtract Equation 1 from Equation 2.
Subtracting Equation 1 from Equation 2 gives:
\[ (A + 13D) - (A + 5D) = 32 - 16 \]
This simplifies to:
\[ 8D = 16 \]

### Step 6: Solve for \( D \).
Dividing both sides by 8:
\[ D = \frac{16}{8} = 2 \]

### Step 7: Substitute \( D \) back into one of the equations to find \( A \).
Using Equation 1:
\[ A + 5D = 16 \]
Substituting \( D = 2 \):
\[ A + 5 \times 2 = 16 \]
\[ A + 10 = 16 \]
\[ A = 16 - 10 = 6 \]

### Step 8: Find the 36th term using the values of \( A \) and \( D \).
Using the formula for the 36th term:
\[ A_{36} = A + (36 - 1)D = A + 35D \]
Substituting \( A = 6 \) and \( D = 2 \):
\[ A_{36} = 6 + 35 \times 2 = 6 + 70 = 76 \]

### Final Answer:
The 36th term of the AP is \( \boxed{76} \).

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