In an increasing sequence of 8 consecutive even integers, the sum of the first 4 integers is 248. What is the sum of all the integers in the sequence?  Choices.
A) 528  B) 568  C) 574  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 =248   or,   4x+12=248   x=236/4= 
<gadget id="calculator">236/4</gadget>
<output>59</output>
59   Thus the answer can now be calculated by either summing up 59+61+63+... 8th term   or   s=n/2(2a+(n-1)d   =8/2(2*59+(7*2))   =4(118+14)=528   528
<result>A</result>
In an increasing sequence of 8 consecutive even integers, the sum of the first 4 integers is 248. What is the sum of all the integers in the sequence?