Prove that for any arithmetic sequence of which the first term is \( a \) and the constant difference is \( d \), the sum to \( n \) terms can be expressed as \( S_{n}=\frac{n}{2}\left(2 a+\left(n^{\prime}-1\right) d\right) \).

To prove the formula for the sum of the first \( n \) terms of an arithmetic sequence, we start with the definition of the sequence. The terms are: - \( a \) (first term) - \( a + d \) (second term) - \( a + 2d \) (third term) - ... - \( a + (n-1)d \) (nth term) The sum \( S_n \) of the first \( n \) terms can be written as: \[ S_n = a + (a + d) + (a + 2d) + \ldots + (a + (n-1)d) \] Now, factor out \( a \) and \( d \): \[ S_n = na + d(0 + 1 + 2 + \ldots + (n-1)) \] The sum of the first \( n-1 \) integers is given by the formula \( \frac{(n-1)n}{2} \). Thus, we have: \[ S_n = na + d\left(\frac{(n-1)n}{2}\right) \] Rearranging gives us: \[ S_n = na + \frac{(n-1)nd}{2} \] We can also express \( S_n \) using another approach by writing the sequence in reverse, and adding both equations: \[ S_n = (a + (n-1)d) + (a + (n-2)d) + \ldots + a \] When we add these equations, each pair sums to the same value: \[ S_n + S_n = n \cdot \text{(first term + last term)} \] Thus: \[ 2S_n = n\left(a + (a + (n-1)d)\right) = n(2a + (n-1)d) \] Dividing by 2 yields: \[ S_n = \frac{n}{2}(2a + (n-1)d) \] This proves the desired formula for the sum of the first \( n \) terms in an arithmetic sequence.