Find the sum to n terms of the A.P., whose kth term is 5k+1.

Explanation

To find the sum to \( n \) terms of the arithmetic progression (A.P.) whose \( k \)-th term is given by \( 5k + 1 \), we will follow these steps:

### Step 1: Identify the first term and common difference
The \( k \)-th term of the A.P. is given as \( a_k = 5k + 1 \).

- To find the first term \( a_1 \), substitute \( k = 1 \):
\[ a_1 = 5(1) + 1 = 5 + 1 = 6 \]

- To find the second term \( a_2 \), substitute \( k = 2 \):
\[ a_2 = 5(2) + 1 = 10 + 1 = 11 \]

- The common difference \( d \) can be calculated as:
\[ d = a_2 - a_1 = 11 - 6 = 5 \]

### Step 2: Use the formula for the sum of the first \( n \) terms of an A.P.
The formula for the sum of the first \( n \) terms \( S_n \) of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 3: Substitute the values into the formula
Now substitute \( a = 6 \) and \( d = 5 \) into the formula:
\[ S_n = \frac{n}{2} \times (2 \times 6 + (n - 1) \times 5) \]

### Step 4: Simplify the expression
Calculate \( 2a \):
\[ 2a = 2 \times 6 = 12 \]

Now substitute this back into the equation:
\[ S_n = \frac{n}{2} \times (12 + (n - 1) \times 5) \]

Distributing \( (n - 1) \times 5 \):
\[ S_n = \frac{n}{2} \times (12 + 5n - 5) \]
\[ S_n = \frac{n}{2} \times (5n + 7) \]

### Step 5: Final expression for \( S_n \)
Thus, the sum of the first \( n \) terms of the A.P. is:
\[ S_n = \frac{n}{2} \times (5n + 7) \]