In an A. P. :
given `a = 7, a _(13)= 35,` find d and `S _(13)`

Answer

To solve the problem step by step, we need to find the common difference \( d \) and the sum of the first 13 terms \( S_{13} \) of the arithmetic progression (A.P.) given that the first term \( a = 7 \) and the 13th term \( a_{13} = 35 \).

### Step 1: Use the formula for the nth term of an A.P.
The formula for the nth term of an A.P. is given by:
\[ a_n = a + (n - 1)d \]
For the 13th term, we have:
\[ a_{13} = a + (13 - 1)d \]
Substituting the known values:
\[ 35 = 7 + (12)d \]

### Step 2: Rearrange the equation to solve for \( d \)
Now, we will rearrange the equation to isolate \( d \):
\[ 35 - 7 = 12d \]
\[ 28 = 12d \]
\[ d = \frac{28}{12} \]
\[ d = \frac{7}{3} \]

### Step 3: Calculate \( S_{13} \) using the sum formula
The formula for the sum of the first \( n \) terms of an A.P. is:
\[ S_n = \frac{n}{2} \left(2a + (n - 1)d\right) \]
For \( S_{13} \):
\[ S_{13} = \frac{13}{2} \left(2 \times 7 + (13 - 1) \times \frac{7}{3}\right) \]

### Step 4: Substitute the values into the sum formula
Calculating the terms inside the parentheses:
\[ S_{13} = \frac{13}{2} \left(14 + 12 \times \frac{7}{3}\right) \]
\[ = \frac{13}{2} \left(14 + 28\right) \]
\[ = \frac{13}{2} \times 42 \]

### Step 5: Simplify the expression
Now, we simplify:
\[ S_{13} = \frac{13 \times 42}{2} \]
\[ = \frac{546}{2} \]
\[ = 273 \]

### Final Result
Thus, the common difference \( d \) is \( \frac{7}{3} \) and the sum of the first 13 terms \( S_{13} \) is \( 273 \).