<li>A sequence of numbers begins 1, 1, 1, 2, 2, 3 and then repeats this pattern forever. What is the sum of the 135th, 136th and 137th numbers in the sequence? </li>
</ol>
<p>(A) 3
(B) 4
(C) 5 <|---- Answer
(D) 6
(E) 7 </p>
<p>I need help with this question. How do you find the 135th term?</p>
<p>Well since 6 is a factor 138 that makes the 138th number equivalent to the 6th number in the sequence. Therefor the 135th, 136th, and the 137th numbers are equivalent to the 3rd, 4th, and 5th numbers respective. 1 + 2 + 2 = 5</p>
<p>Ok 6 is a factor of 138 but so is 2 so then which one do you choose i know you chose 6 but what if a person chooses 2 instead of six and ends up with answer D.</p>
<p>There are 6 numbers. If they repeat, then that means 3 will be repeated everytime the number is a factor of 6. So 6, 12,...120,126,132,138. Because 3 will also be the 138th term, the numbers before in order are the 135,136, and 137. So just add them as X Chu X said. See what im saying? It cant be the second number because the second number is 2, 8, 14...etc..you can caluclate the rest by adding the term +6 (which is the total number of terms before repat).</p>