In this post, we will learn how to derive the formula for the sum of the first \(n\) terms in an arithmetic progression (AP) and see how to apply it through a simple example.
Let \(a_1\) be the first term and \(d\) be the common difference. Then the \(n^{th}\) term of the AP is: \(a_n = a_1 + (n – 1)d \).
The sum of first \(n \) terms is \(S_n = a_1 + a_2 + \dots + a_n \).
\( \begin{align*} S_n &= a_1 + a_2 + a_3 + \dots + a_n \\ &= a_1 + (a_1 + d) + (a_1 + 2d) + \dots + [a_1 + (n – 1)d] \\ &= na_1 + d + 2d + 3d + \dots + (n – 1)d \\ &= na_1 + d(1 + 2 + 3 + \dots + (n – 1)) \\ &= na_1 + d \cdot \frac{(n – 1)n}{2} \quad \left[\because 1 + 2 + \dots + (n – 1) = \frac{(n – 1)n}{2}\right] \\ &= \frac{n}{2}\left[2a_1 + (n – 1)d\right] \dots (1)\\ &= \frac{n}{2}\left[a_1 + \left(a_1 + (n – 1)d\right)\right] \\ &= \frac{n}{2}\left(a_1 + a_n\right) \dots (2) \\ \end{align*} \)You can use either (1) or (2) as the formula for the sum of the first \(n\) terms in an AP.
\( \boxed{S_n = \frac{n}{2}\left[2a_1 + (n – 1)d\right]} \quad \text{(useful when } a_1 \text{ and } d\text{ are known)}\\ \displaystyle \boxed{S_n = \frac{n}{2} (a_1 + a_n)} \quad \text{(useful when the first and last terms are known)}\\ \)Here is an example:
The given series is 6, 9, 12, 15, 18, 21, 24, 27. Find the sum of all terms.
Here \(a_1 = 6\), \(a_n = 27\), \(d = 3 \) and \(n = 8\). We can use either of the two formulas for the sum of this arithmetic progression:
Using the first formula:
\(S_n = \frac{n}{2}\left[2a_1 + (n – 1)d\right] = \frac{8}{2}(2 \times 6 + 7 \times 3) \\= 4 \times (12+21) = 4 \times 33 = 132 \)
Using the second formula:
\(S_n = \frac{n}{2} (a_1 + a_n) = \frac{8}{2}(6+27) = 4 \times 33 = 132 \)