Rebuilding a Smudged Instalment Register
A shopkeeper in Varanasi takes a loan of ₹34,500 from a self-help group to buy a refrigerated display case. Under the group's rules he repays it in monthly instalments that form an arithmetic progression: the first instalment is ₹1,500 and every later instalment is ₹250 more than the one before. The group's register records the twelve instalments and their total, but the last few entries are smudged. The shopkeeper's daughter, a Class 10 student, is asked to rebuild the schedule and to check it against the recorded total. Midway through the year the shopkeeper finds that his season will be better than expected and he wants to finish the loan in 10 instalments instead, still starting at ₹1,500, without borrowing elsewhere. His accountant warns that his cash flow can support an increase of at most ₹300 from one instalment to the next.
1. Rebuild the twelve instalments and check the total against the register's ₹34,500.
Read the worked answer
First instalment a = 1500, common difference d = 250, number of terms n = 12.
Last instalment: a₁₂ = a + 11d = 1500 + 11(250) = 1500 + 2750 = ₹4,250.
Sum: S₁₂ = (n/2)(a + l) = (12/2)(1500 + 4250) = 6 × 5750 = ₹34,500.
Equivalent form: S₁₂ = (12/2)[2(1500) + 11(250)] = 6(3000 + 2750) = 6 × 5750 = ₹34,500.
Schedule: 1500, 1750, 2000, 2250, 2500, 2750, 3000, 3250, 3500, 3750, 4000, 4250.
Add the twelve entries as a check: they total ₹34,500, exactly the figure in the register. The schedule is therefore consistent, and no instalment is negative or out of order - an important check, since an AP with a large d can make later terms unreasonable.
2. After the sixth instalment is paid, how much of the loan is still outstanding?
Read the worked answer
Sum of the first six instalments: S₆ = (6/2)[2(1500) + 5(250)] = 3(3000 + 1250) = 3 × 4250 = ₹12,750.
Outstanding = 34,500 − 12,750 = ₹21,750.
Cross-check by adding the remaining six terms directly: 3000 + 3250 + 3500 + 3750 + 4000 + 4250 = ₹21,750 ✓.
Judgement: it is tempting to say 'half the loan is left after six months' - that would be ₹17,250 and is wrong. Because each instalment is larger than the last, the first six payments cover only about 37% of the loan, not 50%. The shopkeeper is actually behind a straight-line schedule, which matters if he ever wants to settle early.
3. Can he clear the loan in 10 instalments, keeping the first at ₹1,500 and raising each instalment by at most ₹300? Justify with numbers.
Read the worked answer
With a = 1500, n = 10 and d ≤ 300, the largest possible total is obtained at the largest allowed d, d = 300: S₁₀ = (10/2)[2(1500) + 9(300)] = 5(3000 + 2700) = 5 × 5700 = ₹28,500.
Compare with the debt: 28,500 < 34,500. He would be short by 34,500 − 28,500 = ₹6,000.
So under the cash-flow limit the plan is not possible - it is not a matter of choice but of arithmetic.
How much increase would be needed? 5(3000 + 9d) = 34,500 → 3000 + 9d = 6900 → 9d = 3900 → d = ₹433.33 per month. That is about ₹133.33 above his stated limit, so he cannot reach it safely.
Realistic options: keep the 12-instalment schedule, or pay a lump sum of ₹6,000 along with the tenth instalment, or stretch the schedule further rather than steepen it. The key insight is that shortening an AP schedule concentrates the debt into fewer, much larger later terms.