kannn Posted December 11, 2007 Share Posted December 11, 2007 The first three terms of a sequence are 2, 7, and 22. After the first term, each consecutive term can be obtained by multiplying the previous term by 3 and then adding 1. If the sequence continues to be expanded, what will be the sum of the tens digit and the units digit of the thirty-fifth term in the sequence? A. 2 B. 4 C. 7 D. 9 E. 13 Please explain Quote Link to comment Share on other sites More sharing options...
sjd00d Posted December 11, 2007 Share Posted December 11, 2007 B. 4 (not 100% sure though) No quick way, found this out by observing a pattern 02,07,22,67,202,607,1822 .... notice that every odd number ends with 2 and even with 7 and hence our unit digit is 2 for sure. As per tens, observe that the 1st, 5th, .. have it as 0 and 3,7,.. have it as 2 thus the 35th has to be 2. Thus the sum is 4 Quote Link to comment Share on other sites More sharing options...
infinite improbability Posted December 11, 2007 Share Posted December 11, 2007 My answer would be 2 Units digit Between the odds and the evens - it is either 2 or 7 Hence for the 35th term (odd) it is 2 Fifth number in the sequence is 202. Hence tens digit for every fifth term from here onwards would always be 0 Units+Tens = 2 Quote Link to comment Share on other sites More sharing options...
eltonr Posted December 11, 2007 Share Posted December 11, 2007 It is 4 Elton Quote Link to comment Share on other sites More sharing options...
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