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:
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/public/question-images/pasted/1769720849710-mzlz00lufw.png" style="max-width: min(350px, 100%); height: auto; display: block;" alt="Figure 1: Sum of first \(n\) terms of an A.P." class="aligncenter custom-width" /></figure></div>
Produce another identical set of (orange) rods, turn it \(180^\circ\) around and fit it into the existing set of rods: 
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/public/question-images/pasted/1769720976307-yty9pg5sjyl.png" style="max-width: min(400px, 100%); height: auto; display: block;" alt="Figure 2: Fitting another identical set to make a rectangle" class="aligncenter custom-width" /></figure></div>
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, \[
S_n = n \big[ a + (a + (n-1)d) \big] = n (u_1+u_n),
\] 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:
\[
S_n = \frac{n}{2} \big[ 2a + (n-1)d\big] = \frac{n}{2} (u_1 + u_n).
\]

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:

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

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,
\[
S_n = n \big[  a + (a + (n-1)d) \big] = n (u_1+u_n),
\]
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:

Sum of AP formula

Sum of AP formula