Sum of the First n Terms of an Arithmetic Progression
1. Key Property of Arithmetic Progressions
Why does this work? In an AP with first term $a_1$ and common difference $d$:
$$a_p + a_m = (a_1 + (p-1)d) + (a_1 + (m-1)d) = 2a_1 + (p + m - 2)d$$
$$a_k + a_l = (a_1 + (k-1)d) + (a_1 + (l-1)d) = 2a_1 + (k + l - 2)d$$
Since $p + m = k + l$, both expressions are identical. ✓
This is because if a term is $k$ positions from the start and another is $k$ positions from the end, their indices add up to $1 + n$.
2. Deriving the Sum Formula (Gauss's Method)
Let $S_n$ denote the sum of the first $n$ terms of an AP:
Step 1: Write $S_n$ in ascending order:
$S_n = a_1 + a_2 + a_3 + \cdots + a_{n-1} + a_n$
Step 2: Write $S_n$ again in descending (reverse) order:
$S_n = a_n + a_{n-1} + a_{n-2} + \cdots + a_2 + a_1$
Step 3: Add both rows term by term:
$2S_n = (a_1 + a_n) + (a_2 + a_{n-1}) + (a_3 + a_{n-2}) + \cdots + (a_n + a_1)$
Step 4: By the equidistant property, each pair gives $(a_1 + a_n)$, and there are $n$ such pairs:
$2S_n = n \cdot (a_1 + a_n)$
Step 5: Divide both sides by 2:
This is sometimes described as: the sum equals the average of the first and last terms multiplied by the number of terms.