Adding up an AP

"Find 1 + 2 + … + 10 without adding term by term." Write it twice, once backwards:

• 1 + 2 + … + 10
• 10 + 9 + … + 1

Each pair (1 + 10, 2 + 9, …) makes 11, and there are 10 pairs. So 2S = 110 and S = 55.

The same trick works for any n: 2S = n(n + 1), so Sₙ = n(n + 1)/2. In dots, two copies of the staircase 1 + 2 + … + n fit together into an n by (n + 1) rectangle.

Āryabhaṭa's Āryabhaṭīya (Chapter 2, Verse 19) is the first known written statement of this: the average of the first and last terms, times the number of terms.

Sums of a stretch: 25 + 26 + … + 58 = S₅₈ – S₂₄ = 1711 – 300 = 1411. Take away everything up to 24, not up to 25, or you lose the 25.

Triangular numbers are these sums: the nth triangular number is Sₙ. The 10th is 55; the 80th is 3240.

For AP problems, turn the facts into equations in a and d. a + 2d = 12 and a + 49d = 106: subtract to get 47d = 94, so d = 2 and a = 8.