Question

1. the arithmetic mean (average) of 3 consecutive integers is 43. what is the sum of the smallest and largest numbers?

Explanation:

Step1: Recall the formula for the mean of consecutive integers

Let the three consecutive integers be \( n \), \( n + 1 \), \( n + 2 \). The arithmetic mean of these three numbers is given by \( \frac{n+(n + 1)+(n + 2)}{3} \). We know the mean is 43, so we can set up the equation:
\[
\frac{n+(n + 1)+(n + 2)}{3}=43
\]

Step2: Simplify the equation and solve for \( n \)

First, simplify the numerator: \( n+(n + 1)+(n + 2)=3n + 3 \). So the equation becomes:
\[
\frac{3n + 3}{3}=43
\]
Simplify the left - hand side: \( \frac{3n+3}{3}=n + 1 \). So we have \( n + 1=43 \), which means \( n=43 - 1=42 \).

Step3: Find the three consecutive integers

The three consecutive integers are \( n = 42 \), \( n + 1=43 \), \( n + 2 = 44 \).

Step4: Calculate the sum of the smallest and largest numbers

The smallest number is 42 and the largest number is 44. Their sum is \( 42+44 = 86 \).

Answer:

86