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03 Jun 2009, 18:42
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A sequence, a1=64, a2=66, a3=67, an=8+an-3, which of the following is in the sequence?
105
786
966
1025
105
786
966
1025
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03 Jun 2009, 19:13
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vannu
Thats a good question but AC (answer choice) E is missing.
a1 = 64
a2 = 66
a3 = 67
an = 8 + a(n-3)
a3 = 8 + a0
a0 = 67-8 = 59
a4 = a1 + 8 = 64 + 8 = 72
a5 = a2 + 8 = 66 + 8 = 74
a6 = a3 + 8 = 67 + 8 = 75
A. (105 - 64)/8 = k where k must be an integer but k is not.
(105 - 66)/8 = m where m must be an integer but m is not.
(105 - 67)/8 = n where n must be an integer but n is not.
So A is not
B. (786 - 64)/8 = x where x must be an integer but x is not.
(786 - 66)/8 = 720/8 = 90 = y where y must be an integer and y is an integer. Bingo....
(786 - 67)/8 = z where z must be an integer but z is not.
B is.
C. (966 - 64)/8 = p where p must be an integer but p is not.
(966 - 66)/8 = q where q must be an integer but q is not.
(966 - 67)/8 = r where r must be an integer but r is not.
C is not.
D. (1025 - 64)/8 = a where a must be an integer but a is not.
(1025 - 66)/8 = b where b must be an integer but b is not.
(1025 - 67)/8 = c where c must be an integer but c is not.
D is not.
Therefore it is B.
04 Jun 2009, 01:58
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MY logic of approach..
All the numbers in this sequence will remainder 0,2,3 when divided by 8.
So if the remainder is one of the above, its in the sequence..
B gives teh remaider 2.. and hence in the sequence.
All the numbers in this sequence will remainder 0,2,3 when divided by 8.
So if the remainder is one of the above, its in the sequence..
B gives teh remaider 2.. and hence in the sequence.
