What number will come immediately before 63 in the given sequence? 1, 2, 3, 7,…

Concept:

This is an interleaved (alternating) series — two independent numeric chains occupy alternating positions within one visible sequence. To extend such a series, split it by position parity into its two component chains, identify each chain's own arithmetic rule, and then apply that rule to each chain separately.

Application:

1. Split the 8 given terms by position — odd positions (1st, 3rd, 5th, 7th): 1, 3, 7, 15; even positions (2nd, 4th, 6th, 8th): 2, 7, 22, 67.

2. Odd-position chain rule — each term = previous term × 2 + 1: 1×2+1=3, 3×2+1=7, 7×2+1=15.

3. Even-position chain rule — each term = previous term × 3 + 1: 2×3+1=7, 7×3+1=22, 22×3+1=67.

4. Extend the odd-position chain: 9th term = 15×2+1 = 31; 11th term = 31×2+1 = 63.

5. Extend the even-position chain: 10th term = 67×3+1 = 202.

6. So the merged series continues …, 15 (7th), 67 (8th), 31 (9th), 202 (10th), 63 (11th), … — the term immediately before 63 is the 10th term.

Cross-check:

Both chains hold internally all the way through — odd-position chain 1, 3, 7, 15, 31, 63 (each term = previous × 2 + 1); even-position chain 2, 7, 22, 67, 202 (each term = previous × 3 + 1). Since 63 is the 11th (odd-position) term, its immediate predecessor in the merged series is the 10th (even-position) term, 202.