A sequence consists of 16 consecutive even integers written in increasing order. The sum of the first 8 of these even integers is 424. What is the sum of the last 8 of the even integers?  Choose the most appropriate option
A) 488 B) 540 C) 552 D) 568 E) 584
Sum of n even consecutive integers = n(n+1)   Sum of 8 consecutive even integers that start at some point after n = (n+8)(n+9)   Given =&gt; (n+8)(n+9) - n(n+1) = 424   16n +72 = 424 - General equation for sum of consecutive 8 digits where n is where counting starts   Solving we get n = 22 (Where the count starts)   Sum of 8 consecutive integers after n = 22 will be count starting at n1 = 22+8= 
<gadget id="calculator">22+8</gadget>
<output>30</output>
30   So Answer= 16*30 +72= 
<gadget id="calculator">16*30 +72</gadget>
<output>552</output>
552   ANSWER:552
<result>C</result>
A sequence consists of 16 consecutive even integers written in increasing order. The sum of the first 8 of these even integers is 424. What is the sum of the last 8 of the even integers?