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Re: the first 3 terms repeat without end [#permalink]
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150 is also a multiple of 2 and 5. How did you get 3 instead of 2 or 5 in this case?
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Re: the first 3 terms repeat without end [#permalink]
4
no of triplets(since 1,-3,4 keeps repeating) upto 150th place=150/3==>50
there are 50 triplets consisting of digits 1,-3,4
1st triplet=[1,-3,4]
2nd triplets=[1,-3,4]
.
.
.
50th triplets=[1,-3,4]
148th=1
149th=-3
150th=4
but we have t find the sum of [150th+151th+152th+153th+154th]
=4+1+(-3)+4+1
=10-3
=7
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Re: the first 3 terms repeat without end [#permalink]
We can make groups of 3 as the pattern repeats so at the 150th location the cursor is at number (4) in group of [1, –3, 4] so it will move till 154th term such a way that it covers [4,1,-3,4,1 ] thus summing them all up we get 10-3=7 , hence (7)
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Re: the first 3 terms repeat without end [#permalink]
1
gregregre wrote:
150 is also a multiple of 2 and 5. How did you get 3 instead of 2 or 5 in this case?


Because the 3 is the cycle of the sequence, the sequence only ends(where 4 is) in the position of 3' s multiple(ignore other numbers, just 3 because 3 is the cycle). You should just ponder whether the position you wish to find is a multiple of 3, and if it is, then it's where the sequence ends(where 4 is).
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Re: the first 3 terms repeat without end [#permalink]
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