The 3rd and 6th term of an arithmetic progression are 4 and 9, respectively. What is the 12th term?

Options

A. 12
B. 19
C. 9
D. 15

Answer

To find the 12th term of the arithmetic progression (AP) where the 3rd term is 4 and the 6th term is 9, we can follow these steps:

### Step 1: Set Up the Equations
Let \( A \) be the first term and \( D \) be the common difference of the AP. The nth term of an AP can be expressed as:
\[ A_n = A + (n-1)D \]
From the problem, we know:
- The 3rd term \( A_3 = 4 \)
- The 6th term \( A_6 = 9 \)

This gives us two equations:
1. \( A + 2D = 4 \) (for the 3rd term)
2. \( A + 5D = 9 \) (for the 6th term)

### Step 2: Subtract the Equations
Now, we will subtract the first equation from the second:
\[ (A + 5D) - (A + 2D) = 9 - 4 \]
This simplifies to:
\[ 3D = 5 \]

### Step 3: Solve for D
Now, we can solve for \( D \):
\[ D = \frac{5}{3} \]

### Step 4: Substitute D Back to Find A
Next, we substitute \( D \) back into one of the original equations to find \( A \). We can use the first equation:
\[ A + 2D = 4 \]
Substituting \( D \):
\[ A + 2 \left(\frac{5}{3}\right) = 4 \]
This simplifies to:
\[ A + \frac{10}{3} = 4 \]
To isolate \( A \), we convert 4 to a fraction:
\[ A + \frac{10}{3} = \frac{12}{3} \]
Now, subtract \( \frac{10}{3} \) from both sides:
\[ A = \frac{12}{3} - \frac{10}{3} = \frac{2}{3} \]

### Step 5: Find the 12th Term
Now that we have both \( A \) and \( D \), we can find the 12th term \( A_{12} \):
\[ A_{12} = A + 11D \]
Substituting the values of \( A \) and \( D \):
\[ A_{12} = \frac{2}{3} + 11 \left(\frac{5}{3}\right) \]
This simplifies to:
\[ A_{12} = \frac{2}{3} + \frac{55}{3} = \frac{57}{3} \]
Finally, simplifying \( \frac{57}{3} \):
\[ A_{12} = 19 \]

### Conclusion
The 12th term of the arithmetic progression is \( 19 \).

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