If the second and seventh terms of an A.P. are 2 and 22 respectively. Find the sum of first 35 terms:

Options

A. 2310
B. 3210
C. 2130
D. none of these

Answer

To solve the problem step by step, we need to find the first term \( A \) and the common difference \( D \) of the arithmetic progression (A.P.) using the information given about the second and seventh terms. Then, we can calculate the sum of the first 35 terms.

### Step 1: Set up the equations for the terms of the A.P.
The formula for the \( n \)-th term of an A.P. is given by:
\[ T_n = A + (n-1)D \]
From the problem, we know:
- The second term \( T_2 = 2 \)
- The seventh term \( T_7 = 22 \)

Using the formula for the second term:
\[ T_2 = A + (2-1)D = A + D = 2 \quad \text{(Equation 1)} \]

Using the formula for the seventh term:
\[ T_7 = A + (7-1)D = A + 6D = 22 \quad \text{(Equation 2)} \]

### Step 2: Solve the equations simultaneously.
We have the following two equations:
1. \( A + D = 2 \)
2. \( A + 6D = 22 \)

Now, we can subtract Equation 1 from Equation 2:
\[ (A + 6D) - (A + D) = 22 - 2 \]
This simplifies to:
\[ 5D = 20 \]
Dividing both sides by 5 gives:
\[ D = 4 \]

### Step 3: Substitute \( D \) back to find \( A \).
Now that we have \( D \), we can substitute it back into Equation 1:
\[ A + D = 2 \]
Substituting \( D = 4 \):
\[ A + 4 = 2 \]
Solving for \( A \):
\[ A = 2 - 4 = -2 \]

### Step 4: Calculate the sum of the first 35 terms.
The sum \( S_n \) of the first \( n \) terms of an A.P. is given by the formula:
\[ S_n = \frac{n}{2} \times (2A + (n-1)D) \]
For \( n = 35 \):
\[ S_{35} = \frac{35}{2} \times (2(-2) + (35-1) \cdot 4) \]
Calculating inside the parentheses:
\[ 2(-2) = -4 \]
\[ (35-1) \cdot 4 = 34 \cdot 4 = 136 \]
So,
\[ S_{35} = \frac{35}{2} \times (-4 + 136) = \frac{35}{2} \times 132 \]
Calculating further:
\[ S_{35} = \frac{35 \cdot 132}{2} = 35 \cdot 66 = 2310 \]

### Final Answer:
The sum of the first 35 terms of the A.P. is \( 2310 \).

---