Sum of AP formula

Other than a closed formula for the \(n^{\text{th}}\) term of a given arithmetic sequence, it is also useful to find a closed formula for the sum \(S_n\) of its first \(n\) terms. Nothing beats a picture that illustrates the sum of the first \(n\) terms of an arithmetic sequence!

Visualize the finite series \(S=u_1+u_2+u_3+\ldots+u_n\) as the sum of the length of the \(n\) rods (blue) whose lengths form an arithmetic sequence \(u_1,u_2,u_3,\ldots\), where
\[ u_2-u_1 = u_3-u_2 = \ldots = u_n-u_{n-1}=d \]
in the figure below:

Figure 1: Sum of first \(n\) terms of an A.P.

Produce another identical set of (orange) rods, turn it \(180^\circ\) around and fit it into the existing set of rods:

Figure 2: Fitting another identical set to make a rectangle

Thus, the total length of two identical set of rods is \(n (u_1 + u_n ) \), i.e., the total length of one set of rods is
\[ S_n = \frac{n(u_1+u_n)}{2} = \frac{n}{2} \big[ a + (a+(n-1)d) \big] = \frac{n}{2} \big[ 2a + (n-1)d \big]. \]

Mathematically, we can present the above pictorial proof rigorously like this:
\[ \begin{align*} S_n & = a + (a+d) + (a+2d) + \cdots + \big[ a+(n-1)d \big] \\ S_n & = \big[ a + (n-1)d \big] + \big[ a + (n-2)d \big] + \big[ a+(n-3)d\big] + \cdots + a \end{align*} \]

Adding the two equations,
\[ 2S_n = n \big[ a + (a + (n-1)d) \big] = n (u_1+u_n), \]
which implies that the sum \(S_n\) of the first \(n\) terms of an arithmeticc sequence whose first term is \(a\) and common difference \(d\) is given by:

\[ S_n = \frac{n}{2} \big[ 2a + (n-1)d\big] = \frac{n}{2} (u_1 + u_n). \]