Question

(b) trouve 3 nombres entiers consécutifs dont la somme est -147.

Explanation:

Step1: Define the consecutive integers

Let the first integer be \( n \). Then the next two consecutive integers are \( n + 1 \) and \( n + 2 \).

Step2: Set up the equation for the sum

The sum of these three consecutive integers is \( n+(n + 1)+(n + 2) \), and this sum is equal to \(-147\). So we have the equation:
\[
n+(n + 1)+(n + 2)=-147
\]

Step3: Simplify and solve the equation

First, simplify the left - hand side of the equation:
\[

$$\begin{align*} n+(n + 1)+(n + 2)&=-147\\ n + n+1 + n + 2&=-147\\ 3n+3&=-147 \end{align*}$$

\]
Subtract 3 from both sides of the equation:
\[

$$\begin{align*} 3n+3 - 3&=-147-3\\ 3n&=-150 \end{align*}$$

\]
Divide both sides by 3:
\[
n=\frac{-150}{3}=- 50
\]

Step4: Find the other two integers

If \( n=-50 \), then \( n + 1=-50 + 1=-49 \) and \( n + 2=-50+2=-48 \)

Answer:

The three consecutive integers are \(-50\), \(-49\) and \(-48\)