Problem 5 - Entrance Test

Given S_n = n(n+1)(n+2)/6. The nth term t_n = S_n - S_{n-1}. Compute S_{n-1} = (n-1)n(n+1)/6. So t_n = [n(n+1)(n+2) - (n-1)n(n+1)]/6 = n(n+1)[(n+2) - (n-1)]/6 = n(n+1)(3)/6 = n(n+1)/2. So t_n = n(n+1)/2 = (n^2 + n)/2. This is the nth term. The problem states the series is an arithmetico-geometric progression (AGP). An AGP has terms of the form (a + (n-1)d) * r^{n-1}. Let me write t_n = (a + (n-1)d) * r^{n-1}. We have t_n = n(n+1)/2. Let me find the first few terms: t_1 = 1(2)/2 = 1. t_2 = 2(3)/2 = 3. t_3 = 3(4)/2 = 6. t_4 = 4(5)/2 = 10. t_5 = 5(6)/2 = 15. So the series is 1, 3, 6, 10, 15, ... which are triangular numbers. This is not a standard AGP. But the problem says it is an AGP. Let me check if this can be expressed as an AGP. For an AGP, the ratio of consecutive terms is not constant but follows a pattern. Let me compute ratios: t_2/t_1 = 3, t_3/t_2 = 2, t_4/t_3 = 10/6 = 5/3, t_5/t_4 = 15/10 = 3/2. These ratios are 3, 2, 5/3, 3/2,... which are (3/1), (4/2), (5/3), (6/4),... i.e., (n+2)/(n) for the ratio from t_n to t_{n+1}? Let's see: t_2/t_1 = 3 = (1+2)/1? (1+2)/1=3, yes. t_3/t_2 = 2 = (2+2)/2? (2+2)/2=2, yes. t_4/t_3 = 5/3 = (3+2)/3? (3+2)/3=5/3, yes. So t_{n+1}/t_n = (n+2)/n. This means t_{n+1} = t_n * (n+2)/n. This is not a constant ratio, so it's not a GP. For an AGP, the general term is (a + (n-1)d) * r^{n-1}. Let me try to match: t_n = n(n+1)/2. Write as: t_n = (n^2 + n)/2. For large n, t_n ≈ n^2/2, which grows polynomially, not exponentially. An AGP with |r|<1 would converge, but here terms grow. If r>1, terms grow exponentially, but here growth is polynomial. So this cannot be an AGP unless r=1. If r=1, then t_n = a + (n-1)d, which is an AP. But t_n = n(n+1)/2 is quadratic, not linear. So the problem statement seems inconsistent. However, for the sake of the assignment, let me assume the intended series is different. Perhaps S_n = n/2 * (something). Given the options, the common ratio is likely 2/3 (Option B).