In an increasing sequence of 8 consecutive even integers, the sum of the first 4 integers is 212. What is the sum of all the integers in the sequence?
Choices:
A) 252 B) 368 C) 456 D) 586 E) 590
let the first term of the sequence be x   since it is consecutive even integers the terms be we x, x+2,x+4...x+14 (up to 8 terms)   now,   sum of first 4 terms =212   or,   4x+12=212   x=200/4= 
<gadget id="calculator">200/4</gadget>
<output>50</output>
50   Thus the answer can now be calculated by either summing up 50+52+54+... 8th term   or   s=n/2(2a+(n-1)d   =8/2(2*50+(7*2))   =4(100+14)=456   456
<result>C</result>
In an increasing sequence of 8 consecutive even integers, the sum of the first 4 integers is 212. What is the sum of all the integers in the sequence?