The 6th term of an AP is five times the 1st term and the 11th term exceeds twice the 5th term by 3. Find the 8th term of the AP.

Answer

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

### Step 1: Define the terms of the AP
Let:
- \( A \) = first term of the AP
- \( D \) = common difference of the AP

The general formula for the \( n \)-th term of an AP is given by:
\[ A_n = A + (n-1)D \]

### Step 2: Write the equations based on the problem statement
1. The 6th term of the AP is five times the 1st term:
\[ A_6 = 5A \]
Using the formula for the 6th term:
\[ A + 5D = 5A \]
Rearranging gives us:
\[ 5D = 4A \]
This is our **first equation**.

2. The 11th term exceeds twice the 5th term by 3:
\[ A_{11} = 2A_5 + 3 \]
Using the formula for the 11th and 5th terms:
\[ A + 10D = 2(A + 4D) + 3 \]
Simplifying this:
\[ A + 10D = 2A + 8D + 3 \]
Rearranging gives us:
\[ 10D - 8D = 2A - A + 3 \]
\[ 2D = A + 3 \]
This is our **second equation**.

### Step 3: Solve the equations
We have two equations:
1. \( 5D = 4A \) (Equation 1)
2. \( 2D = A + 3 \) (Equation 2)

From Equation 1, we can express \( A \) in terms of \( D \):
\[ A = \frac{5D}{4} \]

Now, substitute \( A \) in Equation 2:
\[ 2D = \frac{5D}{4} + 3 \]

### Step 4: Clear the fraction
To eliminate the fraction, multiply the entire equation by 4:
\[ 8D = 5D + 12 \]

### Step 5: Isolate \( D \)
Subtract \( 5D \) from both sides:
\[ 8D - 5D = 12 \]
\[ 3D = 12 \]
\[ D = 4 \]

### Step 6: Find \( A \)
Now substitute \( D = 4 \) back into the equation for \( A \):
\[ A = \frac{5 \times 4}{4} = 5 \]

### Step 7: Find the 8th term
Now we can find the 8th term \( A_8 \):
\[ A_8 = A + 7D \]
\[ A_8 = 5 + 7 \times 4 \]
\[ A_8 = 5 + 28 \]
\[ A_8 = 33 \]

### Final Answer
The 8th term of the AP is **33**.

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