The sum of a series of terms in A.P. is 128. If the first term is 2 and the last term is 14, find the common difference.

Answer

To solve the problem step by step, we will use the formulas related to an Arithmetic Progression (A.P.).

### Step 1: Identify the given values
- The sum of the series \( S_n = 128 \)
- The first term \( a = 2 \)
- The last term \( l = 14 \)

### Step 2: Use the formula for the sum of an A.P.
The formula for the sum of the first \( n \) terms of an A.P. is given by:

\[ S_n = \frac{n}{2} (a + l) \]

Substituting the known values into the formula:

\[ 128 = \frac{n}{2} (2 + 14) \]

### Step 3: Simplify the equation
Calculate \( a + l \):

\[ 2 + 14 = 16 \]

Now substitute back into the equation:

\[ 128 = \frac{n}{2} \times 16 \]

### Step 4: Solve for \( n \)
Multiply both sides by 2 to eliminate the fraction:

\[ 256 = n \times 16 \]

Now, divide both sides by 16:

\[ n = \frac{256}{16} = 16 \]

### Step 5: Use the formula for the \( n \)-th term of an A.P.
The formula for the \( n \)-th term \( T_n \) of an A.P. is:

\[ T_n = a + (n - 1) \cdot d \]

Where \( d \) is the common difference. Since we know the last term \( T_n = 14 \), we can set up the equation:

\[ 14 = 2 + (16 - 1) \cdot d \]

### Step 6: Simplify and solve for \( d \)
First, simplify the equation:

\[ 14 = 2 + 15d \]

Subtract 2 from both sides:

\[ 12 = 15d \]

Now, divide both sides by 15:

\[ d = \frac{12}{15} = \frac{4}{5} \]

### Final Answer
The common difference \( d \) is \( \frac{4}{5} \).

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