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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  [#permalink]

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New post 20 Dec 2015, 06:09
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Re: In an increasing sequence of 8 consecutive even integers, the sum of  [#permalink]

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New post 20 Dec 2015, 08:14
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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
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Re: In an increasing sequence of 8 consecutive even integers, the sum of  [#permalink]

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New post 20 Dec 2015, 08:19
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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New post 20 Dec 2015, 08:22
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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]

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New post 09 Mar 2016, 22:12
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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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New post 22 May 2017, 05:28
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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
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Re: In an increasing sequence of 8 consecutive even integers, the sum of  [#permalink]

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New post 23 May 2017, 17:15
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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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New post 06 Jun 2017, 00:21
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]

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Re: In an increasing sequence of 8 consecutive even integers, the sum of   [#permalink] 04 Nov 2018, 07:33
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