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09 Feb 2022, 21:41
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B
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Difficulty:
95%
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32% (03:27) correct
68%
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wrong
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Not Attempted Yet
Consider the hexagonal array shown below. Then nth hex number is defined as the total number of dots in the first n layers. The first seven hex numbers are 1, 7, 19, 37, 61, 91 and 127.
What is the 50th hex number?
A. 294
B. 7,351
C. 7,650
D. 7,651
E. 8,751
Attachment:
image0.jpeg [ 121.12 KiB | Viewed 2698 times ]
kungfury42
Joined: 07 Jan 2022
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06 Mar 2022, 14:08
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The answer to this should be B.
The sequence 1, 7, 19, 37, 61, 91, 127 clearly follows an Arithmetic Progression pattern amongst the differences of successive terms. In such a case the general term of the sequence is represented by a quadratic function an²+bn+c.
1st term: a+b+c = 1
2nd term: 4a+2b+c = 7
3rd term: 9a+3b+c = 19
Solving these equations, we get a=3, b=-3 and c=1. We can plug these values back and verify too if we want.
Hence, the 50th term of this series would be a(50)²+b(50)+c or 3(50)²-3(50)+1 = 3(2500)-150+1 = 7500-150+1 = 7350+1 = 7351
Hence, B.
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The sequence 1, 7, 19, 37, 61, 91, 127 clearly follows an Arithmetic Progression pattern amongst the differences of successive terms. In such a case the general term of the sequence is represented by a quadratic function an²+bn+c.
1st term: a+b+c = 1
2nd term: 4a+2b+c = 7
3rd term: 9a+3b+c = 19
Solving these equations, we get a=3, b=-3 and c=1. We can plug these values back and verify too if we want.
Hence, the 50th term of this series would be a(50)²+b(50)+c or 3(50)²-3(50)+1 = 3(2500)-150+1 = 7500-150+1 = 7350+1 = 7351
Hence, B.
Posted from my mobile device
Oppenheimer1945
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Schools: INSEAD '26 (D) ISB '27 (D) IIMB '26 (A) IIMA '26 (A) HEC '26 (D) LBS '26 IESE '27 (D) Fuqua '27 (WL) Tepper '27 (WL)
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27 May 2023, 13:34
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Diff b/w numbers in AP means the general term is a quadratic.
Take ax^2+bx+c
: 3x^2-3x+1 , put x=50: you get 7351
Posted from my mobile device
Take ax^2+bx+c
: 3x^2-3x+1 , put x=50: you get 7351
Posted from my mobile device
