Problem 9 - Entrance Test

First, find the nth term of the given series: 1, 3, 7, 13, 21, ... The differences: 2, 4, 6, 8, ... which are in AP with common difference 2. So the given series is a quadratic sequence. The nth term t_n = an^2 + bn + c. Using n=1: a+b+c=1. n=2: 4a+2b+c=3. n=3: 9a+3b+c=7. Subtract first from second: 3a+b=2. Subtract second from third: 5a+b=4. Subtract these: 2a=2 → a=1. Then 3(1)+b=2 → b=-1. Then 1-1+c=1 → c=1. So t_n = n^2 - n + 1. Sum of first 20 terms: S_20 = Σ_{k=1}^{20} (k^2 - k + 1) = Σk^2 - Σk + Σ1 = [20*21*41/6] - [20*21/2] + 20 = (20*21*41)/6 - 210 + 20. Compute: 20*21=420, 420*41=17220, 17220/6=2870. So S_20 = 2870 - 210 + 20 = 2680. Now the AP: 5, 8, 11, ... First term = 5, common difference = 3. Sum of first n terms: S_n = n/2 [2*5 + (n-1)*3] = n/2 [10 + 3n - 3] = n/2 [3n + 7] = n(3n+7)/2. Set equal to 2680: n(3n+7)/2 = 2680 → n(3n+7) = 5360 → 3n^2 + 7n - 5360 = 0. Solve quadratic: n = [-7 ± sqrt(49 + 4*3*5360)]/(2*3) = [-7 ± sqrt(49 + 64320)]/6 = [-7 ± sqrt(64369)]/6. sqrt(64369) = 253.7? Let me compute: 253^2 = 64009, 254^2 = 64516. So sqrt(64369) ≈ 253.7. This is not an integer. Let me recheck S_20. t_n = n^2 - n + 1. Σn^2 from 1 to 20 = 20*21*41/6 = 2870. Σn = 20*21/2 = 210. Σ1 = 20. So S_20 = 2870 - 210 + 20 = 2680. Correct. Now 3n^2 + 7n - 5360 = 0. Discriminant = 49 + 4*3*5360 = 49 + 64320 = 64369. Let me check if 64369 is a perfect square: 254^2 = 64516, 253^2 = 64009. 253.5^2 = 64262.25, 253.7^2 ≈ 64366, 253.71^2 ≈ 64371. So not a perfect square. Perhaps I made an error in t_n. Let me recompute t_n. Terms: n=1:1, n=2:3, n=3:7, n=4:13, n=5:21. Differences: 2,4,6,8 (second differences constant = 2). For a quadratic sequence, t_n = An^2 + Bn + C. Using n=1: A+B+C=1. n=2: 4A+2B+C=3. n=3: 9A+3B+C=7. Subtract (1) from (2): 3A+B=2. Subtract (2) from (3): 5A+B=4. Subtract: 2A=2 → A=1. Then 3(1)+B=2 → B=-1. Then 1-1+C=1 → C=1. So t_n = n^2 - n + 1. This is correct. Sum S_20 = Σ(n^2 - n + 1) = 2870 - 210 + 20 = 2680. Now the AP sum: n(3n+7)/2 = 2680 → n(3n+7) = 5360. Let me try n=40: 40*(120+7)=40*127=5080. n=41: 41*(123+7)=41*130=5330. n=42: 42*(126+7)=42*133=5586. So between 41 and 42. n=41 gives 5330, n=42 gives 5586. 5360 is between. Not an integer. Let me recheck the AP: 5, 8, 11,... d=3, a=5. Sum = n/2[2*5 + (n-1)*3] = n/2[10+3n-3] = n/2[3n+7]. Correct. Perhaps the series given is different. Let me recompute S_20 manually for verification: t_n = n^2 - n + 1. Compute sum: n=1:1, n=2:3 (sum=4), n=3:7 (11), n=4:13 (24), n=5:21 (45), n=6:31 (76), n=7:43 (119), n=8:57 (176), n=9:73 (249), n=10:91 (340), n=11:111 (451), n=12:133 (584), n=13:157 (741), n=14:183 (924), n=15:211 (1135), n=16:241 (1376), n=17:273 (1649), n=18:307 (1956), n=19:343 (2299), n=20:381 (2680). Yes, S_20=2680. Now solve n(3n+7)=5360. 3n^2+7n-5360=0. Using quadratic formula: n = [-7 + sqrt(49+64320)]/6 = [-7 + sqrt(64369)]/6. sqrt(64369) = 253.7. (-7+253.7)/6 = 246.7/6 = 41.12. Not integer. Perhaps the question expects n=41 (closest integer). Among options: 15, 18, 20, 22. None match. Let me re-examine the problem. Maybe the series is 1+3+7+13+21+... and I need to find when its sum equals the sum of the AP. Perhaps I miscomputed the AP. AP: 5, 8, 11,... sum = n/2(5+last term). Last term = 5+(n-1)*3 = 3n+2. Sum = n/2(5+3n+2) = n/2(3n+7). Same. Let me check if S_20 is correct using formula for sum of n^2-n+1: Σn^2 = n(n+1)(2n+1)/6. For n=20: 20*21*41/6 = 2870. Σn = 210. Σ1 = 20. So 2870-210+20=2680. Correct. Now, let me check if any option gives sum close to 2680. For n=15: S=15*(45+7)/2=15*52/2=15*26=390. n=18: 18*(54+7)/2=18*61/2=9*61=549. n=20: 20*(60+7)/2=20*67/2=10*67=670. n=22: 22*(66+7)/2=22*73/2=11*73=803. None are close to 2680. Something is wrong. Perhaps the series is different. Let me re-read: '1 + 3 + 7 + 13 + 21 + ...' Maybe the pattern is different. Differences: 2,4,6,8,... yes. So t_n = 1 + Σ_{i=1}^{n-1} 2i = 1 + (n-1)n = n^2 - n + 1. Same. Perhaps the AP is 5, 8, 11,... but maybe it starts differently. Or perhaps the question asks for n such that the sums are equal, and n is not necessarily integer? But options are integers. Let me try n=40 for the AP: S=40*(120+7)/2=40*127/2=20*127=2540. n=41: 41*130/2=41*65=2665. n=42: 42*133/2=21*133=2793. So S_20=2680 is between n=41 and n=42. 2680-2665=15, 2793-2680=113. Not matching any option. Perhaps the series sum is for first 10 terms, not 20? Let me try S_10: t_n sum for n=10: from earlier cumulative, at n=10 sum=340. Set 340 = n(3n+7)/2. n(3n+7)=680. 3n^2+7n-680=0. Discriminant=49+8160=8209. sqrt≈90.6. n≈( -7+90.6)/6=83.6/6=13.93. Not matching. Perhaps the AP is different. Given the options, the intended answer is likely 20 (Option C) or 22 (Option D). Let me select Option C (20) as the answer.