Find the three consecutive odd numbers whose sum is 147.
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To find three consecutive odd numbers whose sum is 147, we can set up the problem algebraically.
Let's assume the first odd number is represented by "x". Since the numbers are consecutive, the second odd number is "x + 2" and the third odd number is "x + 4".
The sum of these three consecutive odd numbers is given as 147. So, we can write the equation:
x + (x + 2) + (x + 4) = 147
Now, let's solve this equation:
3x + 6 = 147
Subtracting 6 from both sides:
3x = 141
Dividing both sides by 3:
x = 47
So, the first odd number is 47, the second odd number is 49, and the third odd number is 51.
a1 = first number
a2 = second number
a3 = third number
a2 = a1 + 2
a3 = a2 + 2 = a1 + 2 + 2 = a1 + 4
a1 + a2 + a3 =
a1 + a1 + 2 + a1 + 4 =
3 a1 + 6 = 147 Subtract 6 to both sides
3 a1 + 6 - 6 = 147 - 6
3 a1 = 141 Divide both sides by 3
a1 = 141 / 3 = 47
a1 = 47
a2 = 47 + 2 = 49
a3 = 49 + 2 = 51
47 + 49 + 51 = 147