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. $16$
• B. $18$
• C. $20$
• D. $24$

Answer 1

The correct option is A $16$

Think of the set of eight consecutive integers as follows: $n,(n+1),(n+2),(n+3),(n+4),(n+5),(n+6)$, and $(n+7)$.

First, find the sum of the first four integers:

$n+(n+1)+(n+2)+(n+3)=4n+6$

Then, find the sum of the next four integers:

$(n+4)+(n+5)+(n+6)+(n+7)=4n+22$

The difference between these two partial sums is:

$(4n+22)-(4n+6)=22-6=16$

Another way you could solve this algebraically is to line up the algebraic expressions for each number so that you can subtract one from the other directly:

Sum of the last four integers. $(n+4)+(n+5)+(n+6)+(n+7)$

Less the sum of the first four integers. $\underset{–}{-[n+(n+1)+(n+2)+(n+3\left)\right]}$

$4+4+4+4=16$

Yet another way to see this outcome is to represent the eight consecutive unknowns with eight lines:

----|----

Each of the first four lines can be matched with one of the second four lines, each of which is $4$ greater:

$----|\underset{–}{+4}\underset{–}{+4}\underset{–}{+4}\underset{–}{+4}$

So the sum of the last four numbers is $4\times 4=16$ greater than the sum of the first four.

Finally, you could pick numbers to solve this problem. For example, assume you pick $1,2,3,4,5,6,7$,and $8$. The sum of the first four numbers is $10$. The sum of the last four integers is $26$. Again, the difference is $26-10=16$.