Q1.
Can the same approach be used to find the sum of 1 + 2 + 3 + … + 100?
Answer
Yes — the method does not care how long the list is. Write the sum forwards, write it backwards underneath, and add column by column.
S = 1 + 2 + 3 + … + 98 + 99 + 100
S = 100 + 99 + 98 + … + 3 + 2 + 1
2S = 101 + 101 + 101 + … + 101 (101 added 100 times)
2S = 100 × 101 = 10100
S = 5050
S = 100 + 99 + 98 + … + 3 + 2 + 1
2S = 101 + 101 + 101 + … + 101 (101 added 100 times)
2S = 100 × 101 = 10100
S = 5050
The same answer follows from the formula Sn = n(n + 1)/2 with n = 100: (100 × 101)/2 = 5050.
Why it happens: pairing the two lines works because as you move right the first line grows by 1 while the second shrinks by 1, so every column carries the same total. That total is always (first term + last term) = 1 + 100 = 101, and there are as many columns as there are terms. This is why the trick generalises to any n at all — and it is the argument that produces the formula, rather than merely checking it.
Did you know? the second half of Āryabhaṭa's rule in the Āryabhaṭīya is exactly this: take the average of the first and last terms, then multiply by the number of terms. Here that is (1 + 100)/2 × 100 = 50.5 × 100 = 5050.