The sum of the thirteenth and twenty first term of an A.P. is 34. Find the sum of the first thirty three terms of the progression.

Options

A. 461
B. 561
C. 569
D. None of these

Answer

To solve the problem step by step, we need to find the sum of the first 33 terms of the arithmetic progression (A.P.) given that the sum of the 13th and 21st terms is 34.

### Step 1: Write the formula for the nth term of an A.P.
The nth term of an A.P. can be expressed as:
\[ t_n = a + (n - 1) \cdot d \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Write the expressions for the 13th and 21st terms.
Using the formula from Step 1:
- The 13th term \( t_{13} \) is:
\[ t_{13} = a + (13 - 1) \cdot d = a + 12d \]
- The 21st term \( t_{21} \) is:
\[ t_{21} = a + (21 - 1) \cdot d = a + 20d \]

### Step 3: Set up the equation based on the given information.
According to the problem, the sum of the 13th and 21st terms is 34:
\[ t_{13} + t_{21} = 34 \]
Substituting the expressions from Step 2:
\[ (a + 12d) + (a + 20d) = 34 \]
This simplifies to:
\[ 2a + 32d = 34 \]

### Step 4: Simplify the equation.
Dividing the entire equation by 2:
\[ a + 16d = 17 \]
This is our first equation.

### Step 5: Find the sum of the first 33 terms of the A.P.
The sum \( S_n \) of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \cdot (2a + (n - 1)d) \]
For the first 33 terms (\( n = 33 \)):
\[ S_{33} = \frac{33}{2} \cdot (2a + (33 - 1)d) = \frac{33}{2} \cdot (2a + 32d) \]

### Step 6: Substitute the value of \( 2a + 32d \) from Step 4.
From Step 4, we know:
\[ 2a + 32d = 34 \]
Substituting this into the sum formula:
\[ S_{33} = \frac{33}{2} \cdot 34 \]

### Step 7: Calculate \( S_{33} \).
Calculating:
\[ S_{33} = \frac{33 \cdot 34}{2} = \frac{1122}{2} = 561 \]

Thus, the sum of the first 33 terms of the A.P. is:
\[ \boxed{561} \]