Common terms count in two APs: How many terms are common to the two sequences: AP1 = 1, 3, 5, ... (120 terms) and AP2 = 3, 6, 9, ... (80 terms)?

Introduction / Context:
To count common terms of two APs, characterize the set of numbers belonging to both and count how many such numbers fall within the overlapping numeric range. Here, one AP is odd numbers; the other is multiples of 3.

Given Data / Assumptions:

• AP1: a1 = 1, d1 = 2, n1 = 120 ⇒ last term L1 = 1 + 119*2 = 239.
• AP2: a2 = 3, d2 = 3, n2 = 80 ⇒ last term L2 = 3 + 79*3 = 240.
• Common terms are odd multiples of 3, i.e., numbers congruent to 3 mod 6.

Concept / Approach:
Common sequence C: 3, 9, 15, ... with common difference 6. Count how many are ≤ min(L1, L2) = 239.

Step-by-Step Solution:
C_k = 3 + (k−1)*63 + (k−1)*6 ≤ 239 ⇒ (k−1)*6 ≤ 236 ⇒ k−1 ≤ 39 ⇒ k ≤ 40Hence, there are 40 common terms.

Verification / Alternative check:
The 40th common term is 3 + 39*6 = 237, which is ≤ 239 and also in AP2 (≤ 240).

Why Other Options Are Wrong:
39 undercounts (ignores the 237 term); 41+ would exceed 239.

Common Pitfalls:
Using 240 as the cap (AP1 ends at 239), or taking difference 3 instead of 6 for the intersection.

Final Answer:
40