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The sum of 33 consecutive integers is 1212. Which integers are they?\newline_____\_\_\_\_\_, _____\_\_\_\_\_, _____\_\_\_\_\_

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Q. The sum of 33 consecutive integers is 1212. Which integers are they?\newline_____\_\_\_\_\_, _____\_\_\_\_\_, _____\_\_\_\_\_
  1. Define First Integer: Let xx be the first integer. \newlineConsecutive integers increase by 11 each time. \newlineConsecutive integers: xx, x+1x+1, x+2x+2
  2. Set Up Equation: Set up the equation that represents the sum of 33 consecutive integers equal to 1212.\newlineEquation: x+(x+1)+(x+2)=12x + (x+1) + (x+2) = 12
  3. Simplify Equation: Simplify the left side of the equation x+(x+1)+(x+2)=12x + (x+1) + (x+2) = 12.\newlinex+(x+1)+(x+2)=12x + (x+1) + (x+2) = 12\newline(x+x+x)+(1+2)=12(x + x + x) + (1+2) = 12\newline3x+3=123x + 3 = 12
  4. Isolate Variable Term: Isolate the variable term in 3x+3=123x + 3 = 12.
    3x+3=123x + 3 = 12
    3x+3−3=12−33x + 3 - 3 = 12 - 3
    3x=93x = 9
  5. Solve for x: Solve for x.\newline3x=93x = 9 \newline3x3=93\frac{3x}{3} = \frac{9}{3} \newlinex=3x = 3
  6. Find Consecutive Integers: Find the 33 consecutive integers.\newlineThree consecutive integers: \newlinexx, x+1x+1, x+2x+2 \newline33, 3+13+1, 3+23+2 \newline33, 44, 55

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