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 find the 20th term of the arithmetic progression (A.P.) given that the 6th term is 13 and the 12th term is 25, we can follow these steps:

### Step 1: Write the formulas for the 6th and 12th terms of the A.P.
The nth term of an A.P. is given by the formula:
\[ T_n = a + (n - 1) \cdot d \]
where \( a \) is the first term and \( d \) is the common difference.

For the 6th term:
\[ T_6 = a + (6 - 1) \cdot d = a + 5d \]
Given \( T_6 = 13 \), we have:
\[ a + 5d = 13 \quad \text{(Equation 1)} \]

For the 12th term:
\[ T_{12} = a + (12 - 1) \cdot d = a + 11d \]
Given \( T_{12} = 25 \), we have:
\[ a + 11d = 25 \quad \text{(Equation 2)} \]

### Step 2: Solve the equations
Now we have two equations:
1. \( a + 5d = 13 \) (Equation 1)
2. \( a + 11d = 25 \) (Equation 2)

To eliminate \( a \), we can subtract Equation 1 from Equation 2:
\[ (a + 11d) - (a + 5d) = 25 - 13 \]
This simplifies to:
\[ 11d - 5d = 12 \]
\[ 6d = 12 \]
Dividing both sides by 6 gives:
\[ d = 2 \]

### Step 3: Substitute \( d \) back to find \( a \)
Now that we have \( d \), we can substitute it back into Equation 1 to find \( a \):
\[ a + 5(2) = 13 \]
\[ a + 10 = 13 \]
Subtracting 10 from both sides gives:
\[ a = 3 \]

### Step 4: Find the 20th term
Now that we have both \( a \) and \( d \), we can find the 20th term:
\[ T_{20} = a + (20 - 1) \cdot d \]
Substituting the values of \( a \) and \( d \):
\[ T_{20} = 3 + (19 \cdot 2) \]
\[ T_{20} = 3 + 38 \]
\[ T_{20} = 41 \]

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

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