Determine the sum of first 35 terms of an A.P. if `t_(2)`, = 1 and `t_(7)` , = -22.

Answer

To determine the sum of the first 35 terms of an arithmetic progression (A.P.) given that \( t_2 = 1 \) and \( t_7 = -22 \), we can follow these steps:

### Step 1: Set up the equations for the terms
The nth term of an A.P. is given by the formula:
\[ t_n = A + (n - 1)D \]
where \( A \) is the first term and \( D \) is the common difference.

From the information provided:
- For \( t_2 \):
\[ t_2 = A + (2 - 1)D = A + D = 1 \quad \text{(Equation 1)} \]
- For \( t_7 \):
\[ t_7 = A + (7 - 1)D = A + 6D = -22 \quad \text{(Equation 2)} \]

### Step 2: Solve the equations
Now we have two equations:
1. \( A + D = 1 \)
2. \( A + 6D = -22 \)

We can solve these equations simultaneously. Let's subtract Equation 1 from Equation 2:
\[ (A + 6D) - (A + D) = -22 - 1 \]
This simplifies to:
\[ 5D = -23 \]
Thus, we find:
\[ D = -\frac{23}{5} = -4.6 \]

### Step 3: Substitute to find \( A \)
Now, substitute \( D \) back into Equation 1 to find \( A \):
\[ A + D = 1 \implies A - 4.6 = 1 \]
So,
\[ A = 1 + 4.6 = 5.6 \]

### Step 4: Calculate the sum of the first 35 terms
The sum of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2A + (n - 1)D) \]
For \( n = 35 \):
\[ S_{35} = \frac{35}{2} \times (2 \times 5.6 + (35 - 1)(-4.6)) \]
Calculating inside the parentheses:
\[ 2 \times 5.6 = 11.2 \]
And,
\[ (35 - 1)(-4.6) = 34 \times -4.6 = -156.4 \]
So,
\[ S_{35} = \frac{35}{2} \times (11.2 - 156.4) \]
Calculating the sum inside:
\[ 11.2 - 156.4 = -145.2 \]
Thus,
\[ S_{35} = \frac{35}{2} \times -145.2 = 35 \times -72.6 = -2541 \]

### Final Answer
The sum of the first 35 terms of the A.P. is:
\[ \boxed{-2541} \]