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

Roland2rule 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
OA is c

Let's test 8 consecutive integers

How about: 1, 2, 3, 4, 5, 6, 7, 8

sum of the first four integers = 1 + 2 + 3 + 4 = 10
sum of the last four integers = 5 + 6 + 7 + 8 = 26

Difference = 26 - 10 = 16

Answer: C

Cheers,
Brent

Hi All,

We're told that we have a sequence of 8 consecutive integers. We're asked how much greater the sum of the last four integers is than the sum of the first four integers. This question can be solved rather easily by TESTing VALUES (as Brent has shown). It can also be solved Algebraically:

The 8 consecutive integers can be referred to as:
First 4 numbers: N, N+1, N+2, and N+3
Last 4 numbers: N+4, N+5, N+6 and N+7

Notice how the difference in the first number of each group is 4 (N vs. N+4)... and that each pair of subsequent numbers in each group ALSO differs by 4. Since there are 4 numbers in each group - and the difference is 4 each, the TOTAL difference is (4)(4) = 16.

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

BTGmoderatorRO 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