Write out the sequence:

1, –3, 4, 1, –3, 4

6th term is 4 even though 6 is divisible by 2. We divided by 3 because there are 3 different types of numbers in the list.

For example lets take the following repeating sequence:

2, 5, 7, 8, 9, 2, 5, 7, 8, 9, 2, 5, 7, 8, 9, .....

What is the 15th number? 9 remainder of(15/5)= 0

what is the 22nd number? 5 remainder of (22/5)= 2


So you can vizualize the problem as follows:

remainder 1	remainder 2	remainder 3	remainder 4	remainder 0	remainder 1	remainder 2	remainder 3	remainder 4	remainder 0
2	5	7	8	9	2	5	7	8	9

Does this help?
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Explanation

Examining the repeating pattern, you see that the 3rd term is 4, and every 3rd term after that, in other words, the 6th, 9th, 12th, 15th, and so on, is 4. Since 150 is a multiple of 3, the 150th term is 4. Therefore the 150th to the 154th terms are 4, 1, –3, 4, 1. The sum of these 5 terms is 7, so the correct answer is 7.
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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?

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

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)

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).