Sum of First n Terms of an AP

Today, we’re diving into how to find the sum of the first n terms of an arithmetic progression, or AP. Can anyone remind me what an AP is?

An AP is a sequence of numbers where each term is obtained by adding a constant value to the previous term.

Great! And that constant value is called the common difference, denoted as 'd'. If we want to sum the first n terms, we need to know the first term, 'a', and the common difference. Does anyone know the formula for this sum?

Is it something like S equals n over 2 times the sum of the first and last terms?

Exactly! The formula is $S_n = \frac{n}{2} (a + l)$. We can derive it from the properties of the sequence. Let's explore how.

To derive the sum formula, let’s write down the first two expressions for the sum of terms in an arithmetic sequence. We have S_n = a + (a + d) + (a + 2d) + ... + [a + (n-1)d]. Now, if we write it in reverse order, what do we get?

We get l + (l - d) + (l - 2d) + ... + a.

Right! If we add these two equations, each pair sums to the same value, resulting in $2S = n(2a + (n - 1)d)$. Does anyone remember how we finalize this?

We divide by 2 to get $S_n = \frac{n}{2} [2a + (n - 1)d]$!

Correct! This formula allows us to calculate the sum of the first n terms easily.

Let’s apply this formula to a practical scenario—Shakila's contributions to her daughter's money box over 21 birthdays. Can anyone tell me the first term and common difference here?

The first term, a, is 100, and the common difference, d, is 50, since she increases the amount by that each birthday.

Exactly! Now, applying the sum formula, what would be the total amount by her 21st birthday?

Using $S_n = \frac{n}{2} [2a + (n - 1)d]$, we calculate $S_{21} = \frac{21}{2} [2(100) + (21 - 1)(50)] = \frac{21}{2} [200 + 1000] = \frac{21}{2} [1200] = 12600$.

Well done! This method saves us a lot of time compared to adding each number individually.