AP installments to clear a debt: A debt of ₹3600 is to be repaid in 40 yearly installments forming an AP. After paying 30 installments, one-third of the debt remains unpaid. Find the value of the first installment.

Introduction / Context:
Installments in arithmetic progression allow the use of AP sum formulas. Knowing the total sum (debt) and the partial sum after 30 terms lets us solve for the first term and common difference. Then we can extract the first installment directly.

Given Data / Assumptions:

• Total debt = ₹3600 = S_40.
• Unpaid after 30 installments = one-third of 3600 ⇒ unpaid = 1200, so S_30 = 2400.
• Installments are an AP with first term a and common difference d.

Concept / Approach:
Use S_n = n/2 * [2a + (n − 1)d]. Set up equations for S_30 and S_40 and solve for a and d, then identify a.

Step-by-Step Solution:

Verification / Alternative check:
Check S_40: 20[2*51 + 39*2] = 20[102 + 78] = 20*180 = 3600. Check S_30: 15[102 + 58] = 15*160 = 2400. Consistent.

Why Other Options Are Wrong:

• 55, 53, 49: Do not satisfy both S_30 = 2400 and S_40 = 3600 simultaneously with any constant d.

Common Pitfalls:
Forgetting that the unpaid part fixes S_30, or misusing the AP sum formula (e.g., mixing n − 1 and n). Solve the linear system carefully.

Final Answer: