In an A.P., the `6th` term is `8000` and the `9th` term is `11300`, then which term is 9100?

Options

A. `5^(th)`
B. `6^(th)`
C. `7^(th)`
D. `9^(th)`

Answer

To solve the problem step by step, we need to use the formula for the nth term of an arithmetic progression (A.P.), which is given by:

\[ T_n = a + (n - 1) \cdot d \]

where:
- \( T_n \) is the nth term,
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the term number.

### Step 1: Write the equations for the 6th and 9th terms.

Given:
- The 6th term \( T_6 = 8000 \)
- The 9th term \( T_9 = 11300 \)

Using the formula for the nth term, we can write:

1. For the 6th term:
\[ T_6 = a + (6 - 1) \cdot d = a + 5d = 8000 \quad \text{(Equation 1)} \]

2. For the 9th term:
\[ T_9 = a + (9 - 1) \cdot d = a + 8d = 11300 \quad \text{(Equation 2)} \]

### Step 2: Subtract Equation 1 from Equation 2.

Subtracting Equation 1 from Equation 2 to eliminate \( a \):

\[ (a + 8d) - (a + 5d) = 11300 - 8000 \]

This simplifies to:

\[ 3d = 3300 \]

### Step 3: Solve for \( d \).

Now, divide both sides by 3:

\[ d = \frac{3300}{3} = 1100 \]

### Step 4: Substitute \( d \) back into Equation 1 to find \( a \).

Now that we have \( d \), we can substitute it back into Equation 1 to find \( a \):

\[ a + 5d = 8000 \]

Substituting \( d = 1100 \):

\[ a + 5 \cdot 1100 = 8000 \]

This simplifies to:

\[ a + 5500 = 8000 \]

Subtracting 5500 from both sides gives:

\[ a = 8000 - 5500 = 2500 \]

### Step 5: Find which term is 9100.

Now we need to find \( n \) such that \( T_n = 9100 \):

\[ T_n = a + (n - 1) \cdot d \]

Substituting \( a = 2500 \) and \( d = 1100 \):

\[ 9100 = 2500 + (n - 1) \cdot 1100 \]

### Step 6: Solve for \( n \).

Rearranging the equation:

\[ 9100 - 2500 = (n - 1) \cdot 1100 \]

This simplifies to:

\[ 6600 = (n - 1) \cdot 1100 \]

Now, divide both sides by 1100:

\[ n - 1 = \frac{6600}{1100} = 6 \]

Adding 1 to both sides gives:

\[ n = 7 \]

### Conclusion

Thus, the term that is 9100 is the **7th term**.

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