Complete the series:0, 7, 26, 63, ?

Correct — D.
Rule: n³ − 1 for n = 1, 2, 3, 4 …

1³ − 1 = 0, 2³ − 1 = 7
3³ − 1 = 26, 4³ − 1 = 63
Next: 5³ − 1 = 125 − 1 = 124

Check with gaps: 7, 19, 37 step up by 12, then 18, so the next step is 24.
Next gap = 37 + 24 = 61, and 63 + 61 = 124 → option (d).

Why the others are wrong

• (a)
  100 — 100 repeats the last gap: 63 + 37. The gaps grow each time, 7, 19, 37, so the next one is 61, not 37 again.
• (b)
  125 — 125 is 5³ itself. Every term here is one less than a cube (0 = 1 − 1, 63 = 64 − 1), so the fifth term is 125 − 1.
• (c)
  216 — 216 is 6³. It skips 5³ and drops the −1: after 4³ − 1 = 63 the next term uses n = 5.

When a series grows fast, test it against the squares and cubes. Here every term is one less than a cube: 0, 7, 26, 63 sit just below 1, 8, 27, 64.

The gaps give a second route. First gaps 7, 19, 37 have second gaps 12 and 18, which rise by 6. A series built on n³ has third differences of 6, so the next second gap is 24.

Key facts

• Cubes to know: 1, 8, 27, 64, 125, 216.
• This series is n³ − 1: 0, 7, 26, 63, 124.
• A series built on n³ has constant third differences of 6.

Common traps

• Stopping at 125 because it is a cube, without the −1 every earlier term carries.
• Adding the last gap again (63 + 37 = 100) instead of letting the gaps grow.

No directly related past PYQ was found.