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

Question

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

chetan2u · Answer

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
C

generis · Answer

Method I  - sum consecutive integers

\frac{ (n)}{2}

Sum of first four: 
  \frac{ (4)}{2}   =

2(2x + 3)= 4x + 6

Sum of last four: 
  \frac{ (4)}{2}  = 
2(2x + 11) = 
4x + 22

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

Method II  - use numbers (faster for many)
1, 2, 3, 4, 5, 6, 7, 8

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

Difference: 26 - 10 = 16

Answer C

*  We have  plenty  of time to test another case. 
2, 3, 4, 5, 6, 7, 8, 9
First four, sum: 14
Last four, sum: 30
(30 - 14) = 16

gracie · Answer

sum of first four integers=4x+6
sum of last four integers=4x+22
(4x+22)-(4x+6)=16
C

hawesg · Answer

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

ScottTargetTestPrep · Answer

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

KHow · Answer

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 · Answer

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.

Posted from my mobile device

MathRevolution · Answer

The  5^{th}  number will be 4 more than  1^{st}  number.
The  6^{th}  number will be 4 more than  2^{nd}  number.
The  7^{th}  number will be 4 more than  3^{rd}  number.
The  8^{th}  number will be 4 more than  4^{th}  number.

=> 4 + 4 + 4 + 4 or 4 * 4 = 16

Answer C

bumpbot · Answer

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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

Prep Toolkit

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