In an AP, the sum to 12 terms equals the sum to 18 terms. Find the sum to 30 terms of this AP.

Introduction / Context:
Equating two partial sums in an AP creates a condition on a (first term) and d (common difference). Once a relation is found, we can deduce other sums efficiently.

Given Data / Assumptions:

• S_n = n/2 * (2a + (n−1)d).
• S_12 = S_18.
• Goal: S_30.

Concept / Approach:
Set 12/2(2a+11d) = 18/2(2a+17d) ⇒ 6(2a+11d) = 9(2a+17d). Solve for a in terms of d; then compute S_30.

Step-by-Step Solution:
12a + 66d = 18a + 153d ⇒ 0 = 6a + 87d ⇒ a = −87d/6 = −14.5d.S_30 = 30/2 * (2a + 29d) = 15(2(−14.5d) + 29d) = 15(−29d + 29d) = 0.

Verification / Alternative check:
If S_12 = S_18, then the sum from term 13 to 18 must be zero; symmetrical reasoning extends to S_30 = 0 under the derived a–d relation.

Why Other Options Are Wrong:
1, 2, 3 contradict the established a–d relation; only 0 satisfies the identity for all d ≠ 0 implied by the condition.

Common Pitfalls:
Arithmetic mistakes when distributing 6 and 9; forgetting to halve when computing n/2.

Final Answer:
0