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In a sequence of 8 consecutive integers, how much greater is the sum

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In a sequence of 8 consecutive integers, how much greater is the sum  [#permalink]

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New post 17 Jul 2017, 07:59
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In a sequence of 8 consecutive integers, how much greater is the sum of the last four integers than the sum of the first four integers?

A. 12
B. 14
C. 16
D. 18
E. 20

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Re: In a sequence of 8 consecutive integers, how much greater is the sum  [#permalink]

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New post 17 Jul 2017, 08:04
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susheelh wrote:
In a sequence of 8 consecutive integers, how much greater is the sum of the last four integers than the sum of the first four integers?

A. 12
B. 14
C. 16
D. 18
E. 20




hi..
1st of the first four would be 4 more ta 1st of last four...
similarly 2nd will be 4 more than 2nd of last four..

four terms with each more by 4..
so total sum more by 4*4=16
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In a sequence of 8 consecutive integers, how much greater is the sum  [#permalink]

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New post 17 Jul 2017, 14:43
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susheelh wrote:
In a sequence of 8 consecutive integers, how much greater is the sum of the last four integers than the sum of the first four integers?

A. 12
B. 14
C. 16
D. 18
E. 20

Method I - sum consecutive integers

\(\frac{[(first term) + (last term)](n)}{2}\)

Sum of first four:
\(\frac{[(x) + (x+3)](4)}{2}\) =

2(2x + 3)=

4x + 6

Sum of last four:
\(\frac{[(x+4) + (x+7)](4)}{2}\)=

2(2x +11) =

4x + 22

Difference
4x + 22 - (4x + 6) = 16

Method II - use numbers
1, 2, 3, 4, 5, 6, 7, 8

Sum of first four: 10
Sum of last four: 26

Difference: 26 - 10 = 16

Answer C
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Re: In a sequence of 8 consecutive integers, how much greater is the sum  [#permalink]

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New post 17 Jul 2017, 15:28
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susheelh wrote:
In a sequence of 8 consecutive integers, how much greater is the sum of the last four integers than the sum of the first four integers?

A. 12
B. 14
C. 16
D. 18
E. 20


sum of first four integers=4x+6
sum of last four integers=4x+22
(4x+22)-(4x+6)=16
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Re: In a sequence of 8 consecutive integers, how much greater is the sum  [#permalink]

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New post 15 Mar 2018, 16:20
Or even simpler, realize that each of the 4 numbers is 4 more than its counterpart in the first half and there are 4 of them so the answer is 4*4
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Re: In a sequence of 8 consecutive integers, how much greater is the sum  [#permalink]

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New post 30 Mar 2018, 11:26
1
susheelh wrote:
In a sequence of 8 consecutive integers, how much greater is the sum of the last four integers than the sum of the first four integers?

A. 12
B. 14
C. 16
D. 18
E. 20


We can let the first 4 numbers be 1, 2, 3, and 4, and the last 4 be 5, 6, 7, and 8, thus the difference is:

26 - 10 = 16

Generally, we see that if the first 4 numbers are a, b, c, and d, the next 4 numbers will be (a + 4), (b + 4), (c + 4), and (d + 4). The sum of the first four numbers is (a + b + c+ d), and the sum of the next four numbers is (a + b + c + d + 16). Thus, the sum of the second group of four numbers is 16 greater than the sum of the first four numbers.

Answer: C
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Re: In a sequence of 8 consecutive integers, how much greater is the sum  [#permalink]

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New post 10 Jan 2019, 02:21
Hi Scott,

Thank you for your explanation - I also used this approach to solve it. Will this approach work on the more advanced versions of this problem as well?

Thanks!

ScottTargetTestPrep wrote:
susheelh wrote:
In a sequence of 8 consecutive integers, how much greater is the sum of the last four integers than the sum of the first four integers?

A. 12
B. 14
C. 16
D. 18
E. 20


We can let the first 4 numbers be 1, 2, 3, and 4, and the last 4 be 5, 6, 7, and 8, thus the difference is:

26 - 10 = 16

Generally, we see that if the first 4 numbers are a, b, c, and d, the next 4 numbers will be (a + 4), (b + 4), (c + 4), and (d + 4). The sum of the first four numbers is (a + b + c+ d), and the sum of the next four numbers is (a + b + c + d + 16). Thus, the sum of the second group of four numbers is 16 greater than the sum of the first four numbers.

Answer: C
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Re: In a sequence of 8 consecutive integers, how much greater is the sum  [#permalink]

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New post 10 Jan 2019, 13:11
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Hi susheelh,

It is hard to give a definitive answer to that question; there will be instances where this approach will work, there will be instances where this approach will work with slight modifications, and there will be instances where this approach will not work at all.

For instance, the same method would work if you increased the number of consecutive integers (like 10 consecutive integers instead of 8). You could make it work by modifying the solution if you were looking at the difference between first n terms and last m terms (just represent all the integers as x, x + 1, x + 2 etc. instead of a, b, c, d).

It would also work for consecutive even/odd numbers, just make sure to determine the common difference correctly (in this question, if we had 8 consecutive even integers, we could use the a, b, c, d and a + 8, b + 8, c + 8, d + 8).

The first example that comes to mind in which your method wouldn't work is if, for example, we were looking at the difference between not the sum, but the product of the first so many numbers and the last so many numbers.

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Re: In a sequence of 8 consecutive integers, how much greater is the sum   [#permalink] 10 Jan 2019, 13:11
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