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

Answer

To determine the sum of the first 35 terms of an arithmetic progression (A.P.) given that the second term \( a_2 = 2 \) and the seventh term \( a_7 = 22 \), we can follow these steps:

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

### Step 2: Set up equations based on the given terms.
From the problem, we have:
1. For the second term:
\[ a_2 = a + d = 2 \quad \text{(Equation 1)} \]
2. For the seventh term:
\[ a_7 = a + 6d = 22 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations to find \( a \) and \( d \).
Subtract Equation 1 from Equation 2:
\[ (a + 6d) - (a + d) = 22 - 2 \]
This simplifies to:
\[ 5d = 20 \]
Thus, we find:
\[ d = \frac{20}{5} = 4 \]

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

### Step 4: Use the values of \( a \) and \( d \) to find the sum of the first 35 terms.
The formula for the sum of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
Substituting \( n = 35 \), \( a = -2 \), and \( d = 4 \):
\[ S_{35} = \frac{35}{2} \times (2(-2) + (35 - 1) \cdot 4) \]
Calculating inside the parentheses:
\[ 2(-2) = -4 \]
\[ 35 - 1 = 34 \quad \text{and} \quad 34 \cdot 4 = 136 \]
Thus:
\[ S_{35} = \frac{35}{2} \times (-4 + 136) \]
\[ S_{35} = \frac{35}{2} \times 132 \]
Now calculate:
\[ S_{35} = \frac{35 \cdot 132}{2} = 35 \cdot 66 = 2310 \]

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

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