Determine the sum of first thirty five terms of an arithmetic progression if `t_(2)=2` and `t_(7)=22`.

Answer

To determine the sum of the first thirty-five terms of an arithmetic progression (AP) where \( t_2 = 2 \) and \( t_7 = 22 \), we can follow these steps:

### Step 1: Understand the nth term formula of an AP
The nth term of an arithmetic progression is given by:
\[ t_n = a + (n - 1) \cdot d \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Set up equations using the given terms
From the problem, we know:
- For \( t_2 = 2 \):
\[ t_2 = a + (2 - 1) \cdot d = a + d = 2 \quad \text{(Equation 1)} \]
- For \( t_7 = 22 \):
\[ t_7 = a + (7 - 1) \cdot d = a + 6d = 22 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have two equations:
1. \( a + d = 2 \)
2. \( a + 6d = 22 \)

We can subtract Equation 1 from Equation 2:
\[ (a + 6d) - (a + d) = 22 - 2 \]
This simplifies to:
\[ 5d = 20 \]
Thus, we find:
\[ d = 4 \]

### Step 4: Substitute \( d \) back to find \( a \)
Now substitute \( d = 4 \) back into Equation 1:
\[ a + 4 = 2 \]
So,
\[ a = 2 - 4 = -2 \]

### Step 5: Use the sum formula for the first n terms of an AP
The sum of the first \( n \) terms of an arithmetic progression is given by:
\[ S_n = \frac{n}{2} \cdot (2a + (n - 1) \cdot d) \]
For \( n = 35 \):
\[ S_{35} = \frac{35}{2} \cdot (2(-2) + (35 - 1) \cdot 4) \]

### Step 6: Calculate the sum
Calculating the expression inside the parentheses:
\[ 2(-2) + 34 \cdot 4 = -4 + 136 = 132 \]
Now substituting back into the sum formula:
\[ S_{35} = \frac{35}{2} \cdot 132 \]
Calculating this:
\[ S_{35} = 35 \cdot 66 = 2310 \]

### Final Answer
The sum of the first thirty-five terms of the arithmetic progression is:
\[ \boxed{2310} \]