In an increasing sequence of 8 consecutive even integers, the sum of

Question

In an increasing sequence of 8 consecutive even integers, the sum of the first 4 integers is 268. What is the sum of all the integers in the sequence?

A. 552 
B. 568
C. 574 
D. 586 
E. 590

peachfuzz · Answer

In an increasing sequence of 8 consecutive even integers, the sum of the first 4 integers is 268. What is the sum of all the integers in the sequence?

x+x+2+x+4+x+6=268
x=64

64+66+68+70+72+74+76+78=568

B. 568

CounterSniper · Answer

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 =268
or,
4x+12=268
x=256/4=64
Thus the answer can now be calculated by either summing up 64+66+68+... 8th term 
or
s=n/2(2a+(n-1)d
 =8/2(2*64+(7*2))
 =4(128+14)=568

peachfuzz · Answer

or take the mean*8=568

vardhanindaram · Answer

Another simple way is that as sum of first 4 even is 268,the sum of next of 4 will be 286 + 8(4).This is because difference between corresponding numbers in first and second set of even numbers will always be 8 in this case.Therefore total sum 268+268+32=568.
Answer B

KarishmaB · Answer

The numbers will be of the form:

N, N+2, N+4, N+6, N+8, N+10, N+12, N+14

The sum of first four numbers is 268. The next four numbers are 8 more than each of the previous numbers so their sum will be 268 + 4*8 = 300

Total sum of all 8 numbers = 268 + 300 = 568

ScottTargetTestPrep · Answer

We can label the integers as follows:
 
x = first integer
 
x + 2 = second integer
 
x + 4 = third integer
 
x + 6 = fourth integer
 
x + 8 = fifth integer
 
x + 10 = sixth integer
 
x + 12 = seventh integer
 
x + 14 = eighth integer
 
Since the sum of the first 4 integers is 268:
 
x + x + 2 + x + 4 + x + 6 = 268
 
4x + 12 = 268
 
4x = 256
 
x = 64
 
So, we see that the first integer is 64 and the last integer is 14 + 64 = 78. Since we have an evenly spaced set of integers, the average of the set is (64 + 78)/2 = 142/2 = 71.
 
Since we have 8 integers, the sum is 8 x 71 = 568.
 
Alternative solution:
 
Like the previous solution, we can let x, x + 2, x + 4, x + 6, x + 8, x + 10, x + 12, and x + 14 be the 8 even consecutive integers. Notice that x, x + 2, x + 4, and x + 6 are the first 4 integers and x + 8, x + 10, x + 12, and x + 14 are the last 4 integers. Also notice that each of the last 4 integers is 8 more than each of the first 4 integers, correspondingly; thus, the sum of the last 4 integers is 8 x 4 = 32 more than the sum of the first 4 integers. Since we are given that the first 4 integers sum to 268, the last 4 integers sum to 268 + 32 = 300, and hence the 8 integers sum to 268 + 300 = 568.
 
Answer: B

mahdwen · Answer

You don't wanna lose your time making unnecessary calculations, so maybe look for a pattern to make them easier.
Here, for even numbers sequences, the difference between the first and fifth term is 2x4=8. So the difference in sum between the last four terms and the first four terms is 8x4=32. In other words, the sum of the 8 terms is 268 +   = 568. => Answer B.

Pradd1984 · Answer

For consecutive integers, the median = the average 
268/4 = 67 so first 4 terms are 64,66,68,70 the last 4 are 72 74 76 78
the median number is 71 
71x8 = 568
Answer = B

gracie · Answer

median of first four terms=268/4=67
67 falls halfway between terms two and three
median of all eight terms falls halfway between terms four and five
median for eight terms=67+1+2+1=71
sum of eight terms=8*71=568
B

bumpbot · Answer

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In an increasing sequence of 8 consecutive even integers, the sum of

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In an increasing sequence of 8 consecutive even integers, the sum of

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