Cinema 4

In cinema are 1656 seats and in the last row are 105 seats , in each next row 3 seats less. How many are the total rows in cinema?

Result

n =  23

Solution:

1. a(1) = 105; S(1) = 105
2. a(2) = 102; S(2) = 207
3. a(3) = 99; S(3) = 306
4. a(4) = 96; S(4) = 402
5. a(5) = 93; S(5) = 495
6. a(6) = 90; S(6) = 585
7. a(7) = 87; S(7) = 672
8. a(8) = 84; S(8) = 756
9. a(9) = 81; S(9) = 837
10. a(10) = 78; S(10) = 915
11. a(11) = 75; S(11) = 990
12. a(12) = 72; S(12) = 1062
13. a(13) = 69; S(13) = 1131
14. a(14) = 66; S(14) = 1197
15. a(15) = 63; S(15) = 1260
16. a(16) = 60; S(16) = 1320
17. a(17) = 57; S(17) = 1377
18. a(18) = 54; S(18) = 1431
19. a(19) = 51; S(19) = 1482
20. a(20) = 48; S(20) = 1530
21. a(21) = 45; S(21) = 1575
22. a(22) = 42; S(22) = 1617
23. a(23) = 39; S(23) = 1656

sn=1656 a1=105 d=−3  sn=n(a1+an)/2=n(a1+a1+(n−1)d)/2 1656=n/2(2⋅105−3⋅(n−1)) 3312=n(213−3n) 3312=213n−3n2 3n2−213n+3312=0 n1,2=−b±D2a=213±56256 n1,2=213±756 n1,2=35.5±12.5 n1=48 n2=23 n=23s_n = 1656 \ \\ a_1 = 105 \ \\ d = -3 \ \\ \ \\ s_n = n(a_1+a_n)/2 = n(a_1+a_1+(n-1)d)/2 \ \\ 1656 = n/2 (2 \cdot 105 -3 \cdot (n-1)) \ \\ 3312 = n ( 213 - 3n) \ \\ 3312 = 213 n - 3n^2 \ \\ 3n^2 - 213 n +3312 = 0 \ \\ n_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ 213 \pm \sqrt{ 5625 } }{ 6 } \ \\ n_{1,2} = \dfrac{ 213 \pm 75 }{ 6 } \ \\ n_{1,2} = 35.5 \pm 12.5 \ \\ n_{1} = 48 \ \\ n_{2} = 23 \ \\ n = 23

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