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![If a series is defined as [m]T_n = 3*T_{n-1}+1[/m] and [m]T_n = 2[/m]](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "If a series is defined as [m]T_n = 3*T_{n-1}+1[/m] and [m]T_n = 2[/m]"){kind=icon}
01 Feb 2020, 08:15
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B
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Difficulty:
95%
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Question Stats:
42% (02:37) correct
58%
(02:33)
wrong
based on 81
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If a series is defined as \(T_n = 3*T_{n-1}+1\) and \(T_1= 2\) then find the last two digits of \(T_{50}\)
A) 02
B) 07
C) 22
D) 67
E) None of these
Source: https://www.GMATinsight.com
A) 02
B) 07
C) 22
D) 67
E) None of these
Source: https://www.GMATinsight.com
01 Feb 2020, 09:30
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GMATinsight
\(T_n = 3*T_{n-1}+1\) ... (i)
\(T_n = 2\) should ideally have been: \(T_1 = 02\)
Using (i):
We have: \(T_1\) = 02
n = 2: \(T_2 = 3*T_{1}+1\) = 07
n = 3: \(T_3 = 3*T_{2}+1\) = 22
n = 4: \(T_4 = 3*T_{3}+1\) = 67
n = 5: \(T_5 = 3*T_{4}+1\) = 202 => Last 2 digits is 02
n = 6: \(T_6 = 3*T_{5}+1\) = 607 => Last 2 digits is 07
Thus, we see that the last 2 digits follows a cycle of 4 places: 02, 07, 22 and 67
Thus, the 50th position (i.e. for \(T_{50}\)) will be the 2nd place in the cycle (since the remainder when 50 is divided by 4 is 2) i.e. 07
Answer B
27 Dec 2021, 08:11
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Please edit the coloured part.
