3rd term of an A.P. Is 12 and 10th term is 26, then its 20th term is :

Options

A. 46
B. 52
C. 50
D. 44

Answer

To find the 20th term of the arithmetic progression (A.P.) given that the 3rd term is 12 and the 10th term is 26, we can follow these steps:

### Step 1: Write the formula for the nth term of an A.P.
The nth term of an A.P. can be expressed as:
\[ T_n = A + (n - 1)D \]
where:
- \( T_n \) is the nth term,
- \( A \) is the first term,
- \( D \) is the common difference,
- \( n \) is the term number.

### Step 2: Set up equations using the given terms
From the problem, we know:
- The 3rd term \( T_3 = 12 \):
\[ T_3 = A + (3 - 1)D = A + 2D = 12 \]
(Equation 1)

- The 10th term \( T_{10} = 26 \):
\[ T_{10} = A + (10 - 1)D = A + 9D = 26 \]
(Equation 2)

### Step 3: Solve the equations simultaneously
We have two equations:
1. \( A + 2D = 12 \) (Equation 1)
2. \( A + 9D = 26 \) (Equation 2)

To eliminate \( A \), we can subtract Equation 1 from Equation 2:
\[ (A + 9D) - (A + 2D) = 26 - 12 \]
This simplifies to:
\[ 7D = 14 \]
Now, solve for \( D \):
\[ D = \frac{14}{7} = 2 \]

### Step 4: Substitute \( D \) back to find \( A \)
Now that we have \( D \), we can substitute it back into Equation 1 to find \( A \):
\[ A + 2(2) = 12 \]
\[ A + 4 = 12 \]
\[ A = 12 - 4 = 8 \]

### Step 5: Find the 20th term
Now we can find the 20th term \( T_{20} \):
\[ T_{20} = A + (20 - 1)D = A + 19D \]
Substituting the values of \( A \) and \( D \):
\[ T_{20} = 8 + 19(2) = 8 + 38 = 46 \]

### Final Answer
The 20th term of the A.P. is **46**.