Simran's Monthly Savings Scheme: An Arithmetic Progression Decision
Simran, a Class X student in a PSEB-affiliated school in Bathinda, joins her school's monthly savings scheme to buy a sewing machine for her mother's tailoring work. She deposits ₹500 in the first month and increases the deposit by ₹75 every month thereafter, so the deposits form an arithmetic progression. She plans to run the scheme for two years, that is 24 months. The machine she has chosen costs ₹35,000, and the shop will hold that price for her only if she pays the full amount within 25 months. Her father has said plainly that he will add nothing to the fund and that she must not borrow. Before signing the scheme register, Simran wants three answers: the first month in which her own monthly deposit reaches ₹2,000, the total she will have saved after 24 months, and whether she can reach ₹35,000 by herself - and if so, in which month.
1. In which month does Simran's monthly deposit first reach ₹2,000?
Read the worked answer
Deposits form an AP with first term a = 500 and common difference d = 75.
aₙ = a + (n − 1)d = 500 + (n − 1) × 75
Require 500 + 75(n − 1) ≥ 2000 75(n − 1) ≥ 1500 n − 1 ≥ 20 n ≥ 21
Check the boundary rather than trusting the inequality alone: Month 20: 500 + 19 × 75 = 500 + 1425 = ₹1,925 (still short) Month 21: 500 + 20 × 75 = 500 + 1500 = ₹2,000 exactly
So the 21st month is the first month in which the deposit reaches ₹2,000, and it hits the figure exactly, not above it.
2. What is the total saved after 24 months, and how far is she from the price of the machine?
Read the worked answer
Use Sₙ = (n/2)[2a + (n − 1)d] with n = 24, a = 500, d = 75.
S₂₄ = (24/2)[2(500) + 23(75)] = 12[1000 + 1725] = 12 × 2725 = ₹32,700
Price of the machine = ₹35,000. Shortfall = 35,000 − 32,700 = ₹2,300.
So at the end of the planned two years she is ₹2,300 short and cannot buy the machine on the 24-month plan.
3. Can she reach ₹35,000 on her own, and in which month? Show the inequality and check the boundary.
Read the worked answer
Sₙ = (n/2)[2(500) + (n − 1)75] = (n/2)[1000 + 75n − 75] = (n/2)[925 + 75n]
Require (n/2)(925 + 75n) ≥ 35,000 75n² + 925n ≥ 70,000 Divide throughout by 25: 3n² + 37n − 2800 ≥ 0
Factorise: (n − 25)(3n + 112) = 3n² + 112n − 75n − 2800 = 3n² + 37n − 2800 ✓
Roots: n = 25 and n = −112/3. A month number must be a positive integer, so reject −112/3, leaving n ≥ 25.
Boundary check: Month 25 deposit = 500 + 24 × 75 = 500 + 1800 = ₹2,300 S₂₅ = S₂₄ + 2300 = 32,700 + 2,300 = ₹35,000 exactly
Her month-25 deposit is precisely the ₹2,300 she was short, so she reaches the target exactly in month 25 - inside the shop's 25-month deadline, but with nothing to spare. She should get the price confirmed in writing, because any rise (or a single missed month) puts the machine out of reach and she has ruled out borrowing.