Question

There are three consecutive integers such that twice the sum of the smallest and largest is eight more than twice the largest. Find the smallest integer in the sequence.

a) 3
b) 4
c) 5
d) 6

Answer 1

The smallest integer in the sequence is found by setting up an equation from the given conditions and solving for x. The equation 2(x + (x + 2)) = 2(x + 2) + 8 simplifies to x = 4, thus the smallest integer is 4.

Step-by-step explanation:

The student question involves finding the smallest integer in a sequence of three consecutive integers where twice the sum of the smallest and largest is eight more than twice the largest. If we let x be the smallest integer, then x+1 and x+2 are the second and the largest integers, respectively.

According to the problem, we can write the following equation: 2(x + (x + 2)) = 2(x+2) + 8. Solving for x gives us:

2(2x + 2) = 2x + 4 + 8

4x + 4 = 2x + 12

4x - 2x = 12 - 4

2x = 8

x = 4

Therefore, the smallest integer in the sequence is 4, which corresponds to option b).