If ` 6^(th) and 12^(th)` term of an A.P. are 13 and 25
respectively , then its ` 20^(th)` term is

Options

A. 37
B. 39
C. 41
D. 43

Answer

To solve the problem, we need to find the 20th term of an arithmetic progression (A.P.) given that the 6th term is 13 and the 12th term is 25. We can use the general formula for the nth term of an A.P., which is:

\[ T_n = A + (n - 1)D \]

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 12th terms.

For the 6th term (\( T_6 \)):
\[ T_6 = A + (6 - 1)D = A + 5D = 13 \]
This is our **Equation 1**.

For the 12th term (\( T_{12} \)):
\[ T_{12} = A + (12 - 1)D = A + 11D = 25 \]
This is our **Equation 2**.

### Step 2: Set up the equations.

From **Equation 1**:
\[ A + 5D = 13 \]

From **Equation 2**:
\[ A + 11D = 25 \]

### Step 3: Subtract Equation 1 from Equation 2.

Subtracting **Equation 1** from **Equation 2** gives:
\[ (A + 11D) - (A + 5D) = 25 - 13 \]
This simplifies to:
\[ 11D - 5D = 12 \]
\[ 6D = 12 \]

### Step 4: Solve for D.

Dividing both sides by 6:
\[ D = \frac{12}{6} = 2 \]

### Step 5: Substitute D back into Equation 1 to find A.

Now substitute \( D = 2 \) back into **Equation 1**:
\[ A + 5(2) = 13 \]
\[ A + 10 = 13 \]
Subtracting 10 from both sides:
\[ A = 13 - 10 = 3 \]

### Step 6: Find the 20th term.

Now that we have \( A = 3 \) and \( D = 2 \), we can find the 20th term (\( T_{20} \)):
\[ T_{20} = A + (20 - 1)D \]
\[ T_{20} = 3 + 19(2) \]
\[ T_{20} = 3 + 38 = 41 \]

### Final Answer:
The 20th term of the A.P. is **41**.

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