[Algebra]
Problem: The average of three consecutive multiples of 3 is $a$.

The average of four consecutive multiples of 4 is $a+27$.

The average of the smallest and largest of these seven integers is 42 .

Determine the value of $a$.

Solution: Since the average of three consecutive multiples of 3 is $a$, then $a$ is the middle of these three integers, so the integers are $a-3, a, a+3$.

Since the average of four consecutive multiples of 4 is $a+27$, then $a+27$ is halfway in between the second and third of these multiples (which differ by 4), so the second and third of the multiples are $(a+27)-2=a+25$ and $(a+27)+2=a+29$, so the four integers are $a+21, a+25, a+29, a+33$.

(We have used in these two statements the fact that if a list contains an odd number of integers, then there is a middle integer in the list, and if the list contains an even number



of integers, then the "middle" integer is between two integers from the list.)

The smallest of these seven integers is $a-3$ and the largest is $a+33$.

The average of these two integers is $\frac{1}{2}(a-3+a+33)=\frac{1}{2}(2 a+30)=a+15$.

Since $a+15=42$, then $a=27$.

Final answer: 27