Two savings plans: when does doubling overtake a fixed top-up?
A self-help group in Madurai compares two monthly savings plans for a 10-month equipment fund. Plan A: the first deposit is ₹500, and each later deposit is ₹500 more than the previous one, so the deposits are 500, 1000, 1500, and so on. Plan B: the first deposit is ₹200, and each later deposit is double the previous one, so the deposits are 200, 400, 800, and so on. The treasurer likes Plan B because the individual deposits grow faster, but another member argues that Plan A's larger starting amount may keep its total ahead for several months. They need a month-by-month comparison, not just the 10th deposit. They also want to know the smallest month in which Plan A's deposit is a multiple of 7, using the general term rather than listing, and they want the 10-month totals checked against the claim that each total is a multiple of ₹100.
1. Find the total saved under each plan after n months. Determine the smallest n for which Plan B's total first exceeds Plan A's total.
Read the worked answer
Plan A is an arithmetic progression with first term 500 and common difference 500. Its total is S_A(n) = n/2 [2(500) + (n - 1)500] = 250n(n + 1). Plan B is a geometric progression with first term 200 and common ratio 2. Its total is S_B(n) = 200(2^n - 1)/(2 - 1) = 200(2^n - 1). Check early months: n=1: A=500, B=200; n=2: A=1500, B=600; n=3: A=3000, B=1400; n=4: A=5000, B=3000; n=5: A=7500, B=6200; n=6: A=10500, B=12600. So Plan B first exceeds Plan A at n=6 months.
2. Find the 10th deposit and the 10-month total under each plan. Verify whether each 10-month total is a multiple of ₹100.
Read the worked answer
Plan A: 10th deposit = 500 + 9(500) = 5000. S_A(10) = 250 x 10 x 11 = 27500, which is 275 x 100, so it is exactly a multiple of ₹100. Plan B: 10th deposit = 200 x 2^9 = 200 x 512 = 102400. S_B(10) = 200(2^10 - 1) = 200(1024 - 1) = 200 x 1023 = 204600, which is 2046 x 100. Both totals are multiples of ₹100, so no remainder is left when packed in ₹100 envelopes.
3. Plan A's monthly deposits are 500, 1000, 1500, and so on. Find the smallest month whose deposit is exactly divisible by 7. Use modular arithmetic, not a list.
Read the worked answer
The nth deposit is a_n = 500n. Modulo 7, 500 = 7 x 71 + 3, so 500n = 3n (mod 7). We need 3n = 0 (mod 7). Since 3 and 7 are coprime, 3 has an inverse modulo 7, so n must be a multiple of 7. The smallest positive n is 7. The 7th deposit is 500 x 7 = 3500, and 3500/7 = 500. Thus the first such month is month 7, with deposit ₹3500.