The problem asks to show that the sum of an arithmetic progression (AP) is given by the formula $S_n = \frac{n}{2}[a + l]$, where $S_n$ is the sum of the first $n$ terms, $a$ is the first term, and $l$ is the last term.

1. Problem Description

The problem asks to show that the sum of an arithmetic progression (AP) is given by the formula $S_n = \frac{n}{2}[a + l]$, where $S_n$ is the sum of the first $n$ terms, $a$ is the first term, and $l$ is the last term.

2. Solution Steps

Let the arithmetic progression be $a, a+d, a+2d, ..., a+(n-1)d$, where $a$ is the first term and $d$ is the common difference.

The last term, $l$, can be expressed as $l = a + (n-1)d$.

The sum of the first $n$ terms of an arithmetic progression can be written as:

$S_n = a + (a+d) + (a+2d) + ... + (l-2d) + (l-d) + l$

We can also write the sum in reverse order:

$S_n = l + (l-d) + (l-2d) + ... + (a+2d) + (a+d) + a$

Adding these two equations term by term, we get:

$2S_n = (a+l) + (a+d+l-d) + (a+2d+l-2d) + ... + (l-2d+a+2d) + (l-d+a+d) + (l+a)$

$2S_n = (a+l) + (a+l) + (a+l) + ... + (a+l) + (a+l) + (a+l)$

Since there are $n$ terms in the arithmetic progression, we have $n$ terms of $(a+l)$ on the right-hand side.

$2S_n = n(a+l)$

Dividing both sides by 2, we get:

$S_n = \frac{n}{2}(a+l)$

3. Final Answer

The sum of an arithmetic progression is given by $S_n = \frac{n}{2}(a+l)$.