Chapter 2.6, Problem 47HP

To determine

what is the sum of the last six integers?

Answer to Problem 47HP

Correct answer $\boxed{J}$

Explanation of Solution

Given information:

Twelve consecutive integers are arranged in order from least to greatest. If the sum of the first six integers is $381$ , what is the sum of the last six integers?

F $345$

G $381$

H $387$

J $417$

Calculation:

The integers are consecutive. Therefore, each integer is obtained by adding $1$ to the immediately preceding integer.

So, if $n+0$ is the first integer, then $n+1$ the second integer and the third integer is $n+1+1$ or $n+2$ , and so on. The first six of the integers are:

  $n+0$

  $n+1$

  $n+2$

  $n+3$

  $n+5$

The sum of these $6$ numbers is $6n+\left(0+1+2+3+4+5\right)$

This implies that;

  $6n+15$

Now, the problem states that the sum of the first $6$ numbers is $381$ .

So set up the equation:

  $6n+15=381$

To solve for $n$ ;

  $6n=381-15$

  $6n=366$

  $n=\frac{366}{6}$

  $n=61$

It is known that the first number is $61$ . Therefore, because they are consecutive, the first $6$

numbers are $61,62,63,64,65,66.$ That means the seventh through twelfth numbers are

  $67+68+69+70+71+72=417$

Hence, the correct answer is $\boxed{J}$