Grades 11-12 Video Solutions 2011-11&12 Video Solutions 2011 problem22

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Question number 22. Consider the arithmetic sequences beginning with 5, 20, 35, and so on, and also 35, 61, 87. How many different arithmetic sequences with positive terms contain  all the terms of both these sequences? Let's obtain the general term of each sequence. In the first case, we have a sequence beginning  with 5 with a common difference of 15, so we add multiples of 15 to obtain more terms. The sequence beginning with 5, 20, 35 continues on to 50, 65, 80, and so on. And our second  sequence beginning with 35 has a common difference of 26, so we add multiples of 26, and 35, 61, 87 is followed by 113, and then 139, 165, and so on.  Then looking at this, we notice that both sets of numbers have to be contained in a single sequence. 139 here is sort of special because later on, 140 appears in the first  sequence. 140 is divisible by 15 with remainder 5. So what we notice is that the sequences we are looking for, they have to contain consecutive terms, consecutive integers, that is.  and so with that observation we know that they have the form we will start  with some initial term let's call it a and then just add on integers starting with zero all the way up to infinity so here the number a is the initial term  now we recall that our sequences have to be sequences of positive numbers positive terms only and the choices for a are therefore a can be equal to one  two three four or five if we start with a number greater than five then we will be missing our initial term here of five therefore we have five choices for the  sequences each beginning with either one two three four or five and containing all the consecutive integers thereafter so the answer here is C

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