In an A.P.,8, 11, 14, .......... find `S_(n) – S_(n – 1)`

Answer

To solve the problem of finding \( S_n - S_{n-1} \) for the arithmetic progression (A.P.) given by the terms 8, 11, 14, ..., we can follow these steps:

### Step 1: Identify the first term and common difference
The first term \( a \) of the A.P. is 8, and the common difference \( d \) can be calculated as:
\[ d = 11 - 8 = 3 \]

### Step 2: Write the formula for the sum of the first \( n \) terms, \( S_n \)
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 the values of \( a \) and \( d \):
\[ S_n = \frac{n}{2} \times (2 \times 8 + (n-1) \times 3) \]
\[ S_n = \frac{n}{2} \times (16 + 3n - 3) \]
\[ S_n = \frac{n}{2} \times (3n + 13) \]

### Step 3: Write the formula for \( S_{n-1} \)
Now, we need to find \( S_{n-1} \):
\[ S_{n-1} = \frac{n-1}{2} \times (2a + (n-2)d) \]
Substituting the values:
\[ S_{n-1} = \frac{n-1}{2} \times (16 + (n-2) \times 3) \]
\[ S_{n-1} = \frac{n-1}{2} \times (16 + 3n - 6) \]
\[ S_{n-1} = \frac{n-1}{2} \times (3n + 10) \]

### Step 4: Calculate \( S_n - S_{n-1} \)
Now, we can find \( S_n - S_{n-1} \):
\[ S_n - S_{n-1} = \left( \frac{n}{2} \times (3n + 13) \right) - \left( \frac{n-1}{2} \times (3n + 10) \right) \]
Factor out \( \frac{1}{2} \):
\[ S_n - S_{n-1} = \frac{1}{2} \left( n(3n + 13) - (n-1)(3n + 10) \right) \]
Now, expand both terms:
\[ = \frac{1}{2} \left( 3n^2 + 13n - (3n^2 + 10n - 3n - 10) \right) \]
\[ = \frac{1}{2} \left( 3n^2 + 13n - 3n^2 - 10n + 3n + 10 \right) \]
Combine like terms:
\[ = \frac{1}{2} \left( 6n + 10 \right) \]
\[ = 3n + 5 \]

### Final Answer
Thus, the result is:
\[ S_n - S_{n-1} = 3n + 5 \]

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