Arithmetic progression (AP) – compute the 12th term: If five times the fifth term of an AP equals seven times the seventh term, determine the twelfth term.

Introduction / Context:
Relating two AP terms gives a linear equation in the first term and common difference. Once the relation is solved, the requested term follows by substitution.

Given Data / Assumptions:

• a_n = a + (n − 1)d.
• 5(a + 4d) = 7(a + 6d).

Concept / Approach:
Expand and collect terms to express a in terms of d, then compute a_12 = a + 11d.

Step-by-Step Solution:
5a + 20d = 7a + 42d ⇒ 0 = 2a + 22d ⇒ a = −11d.a_12 = a + 11d = (−11d) + 11d = 0.

Verification / Alternative check:
If a = −11 and d = 1, then a_5 = −7; a_7 = −5; 5(−7) = −35; 7(−5) = −35 (condition holds), and a_12 = 0.

Why Other Options Are Wrong:
Nonzero values contradict the derived identity a = −11d which forces a_12 = 0 for all d.

Common Pitfalls:
Plugging n instead of n − 1; arithmetic slips when moving terms across the equality.

Final Answer:
0