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Re: In an increasing sequence of 8 consecutive even integers, the sum of [#permalink]
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Bunuel wrote:
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


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
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Re: In an increasing sequence of 8 consecutive even integers, the sum of [#permalink]
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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
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Re: In an increasing sequence of 8 consecutive even integers, the sum of [#permalink]
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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
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Re: In an increasing sequence of 8 consecutive even integers, the sum of [#permalink]
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
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Re: In an increasing sequence of 8 consecutive even integers, the sum of [#permalink]
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peachfuzz wrote:
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


or take the mean*8=568
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Re: In an increasing sequence of 8 consecutive even integers, the sum of [#permalink]
Bunuel wrote:
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


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 + [268 + 32] = 568. => Answer B.
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Re: In an increasing sequence of 8 consecutive even integers, the sum of [#permalink]
Bunuel wrote:
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


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
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Re: In an increasing sequence of 8 consecutive even integers, the sum of [#permalink]
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Re: In an increasing sequence of 8 consecutive even integers, the sum of [#permalink]
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