An A.P consists of 50 terms of which `3^(rd)` term is 12 and the last term is 106. Find the `29^(th)` term .

Options

A. 62
B. 64
C. 29
D. 94

Answer

To find the 29th term of the arithmetic progression (A.P.) given that the 3rd term is 12 and the last term (50th term) is 106, we can follow these steps:

### Step 1: Define the terms of the A.P.
In an A.P., the nth term can be expressed as:
\[ T_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Write equations based on the given information.
From the problem, we know:
- The 3rd term \( T_3 = 12 \):
\[ T_3 = a + 2d = 12 \quad \text{(Equation 1)} \]
- The 50th term \( T_{50} = 106 \):
\[ T_{50} = a + 49d = 106 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations simultaneously.
We have the two equations:
1. \( a + 2d = 12 \)
2. \( a + 49d = 106 \)

To eliminate \( a \), we can subtract Equation 1 from Equation 2:
\[ (a + 49d) - (a + 2d) = 106 - 12 \]
This simplifies to:
\[ 49d - 2d = 94 \]
\[ 47d = 94 \]
Now, divide both sides by 47:
\[ d = \frac{94}{47} = 2 \]

### Step 4: Substitute \( d \) back to find \( a \).
Now that we have \( d = 2 \), we can substitute it back into Equation 1 to find \( a \):
\[ a + 2(2) = 12 \]
\[ a + 4 = 12 \]
Subtract 4 from both sides:
\[ a = 12 - 4 = 8 \]

### Step 5: Find the 29th term.
Now that we have both \( a \) and \( d \), we can find the 29th term \( T_{29} \):
\[ T_{29} = a + (29-1)d \]
Substituting the values of \( a \) and \( d \):
\[ T_{29} = 8 + (28)(2) \]
\[ T_{29} = 8 + 56 \]
\[ T_{29} = 64 \]

### Final Answer:
The 29th term of the A.P. is **64**.

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