Question

The sum of three consecutive odd numbers is 249. Find the numbers.

Answer 1

Hint: Here we assume three consecutive odd numbers as \[(2n + 1),(2n + 3),(2n + 5)\]. Add the three numbers and equate the sum to 249 to find the value of n. Substitute back the value of n in the numbers.

Complete step-by-step answer:
We have to take three consecutive natural numbers.
Let the three consecutive odd numbers be \[(2n + 1),(2n + 3),(2n + 5)\]
We are given the sum of three consecutive odd numbers is 249
Add the three numbers:
\[ \Rightarrow (2n + 1) + (2n + 3) + (2n + 5) = 249\]
Add like values in LHS
\[ \Rightarrow 6n + 9 = 249\]
Shift the constant values to RHS of the equation.
\[ \Rightarrow 6n = 249 - 9\]
\[ \Rightarrow 6n = 240\]
Divide both sides of the equation by 6
\[ \Rightarrow \dfrac{{6n}}{6} = \dfrac{{240}}{6}\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow n = 40\]
Now we substitute the value of n in each \[(2n + 1),(2n + 3),(2n + 5)\]to obtain the three numbers.
Put\[n = 40\]in \[(2n + 1)\]
\[ \Rightarrow (2n + 1) = (2 \times 40 + 1)\]
\[ \Rightarrow (2n + 1) = 80 + 1\]
\[ \Rightarrow (2n + 1) = 81\]
So, the first odd number is 81.
Put\[n = 40\] in \[(2n + 3)\]
\[ \Rightarrow (2n + 3) = (2 \times 40 + 3)\]
\[ \Rightarrow (2n + 3) = 80 + 3\]
\[ \Rightarrow (2n + 3) = 83\]
So, the second odd number is 83.
Put\[n = 40\]in \[(2n + 5)\]
\[ \Rightarrow (2n + 5) = (2 \times 40 + 5)\]
\[ \Rightarrow (2n + 5) = 80 + 5\]
\[ \Rightarrow (2n + 5) = 85\]
So, the third odd number is 85.

Therefore, the three consecutive odd numbers are 81, 83 and 85.

Note: Students are likely to make the mistake of not changing the sign of a value when shifting the value from one side of the equation to the other side of the equation, keep in mind sign changes from positive to negative and vice-versa when a value is shifted.
Alternate Method:
We can take three consecutive alternate numbers as \[n + 1,n + 3,n + 5\]
Then the sum of three consecutive odd numbers is 249
\[ \Rightarrow n + 1 + n + 3 + n + 5 = 249\]
Add like values in LHS
\[ \Rightarrow 3n + 9 = 249\]
Shift the constant values to RHS of the equation.
\[ \Rightarrow 3n = 249 - 9\]
\[ \Rightarrow 3n = 240\]
Divide both sides of the equation by 3
\[ \Rightarrow \dfrac{{3n}}{3} = \dfrac{{240}}{3}\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow n = 80\]
Substitute the value of n in \[n + 1,n + 3,n + 5\]
\[ \Rightarrow n + 1 = 80 + 1 = 81\]
\[ \Rightarrow n + 3 = 80 + 3 = 83\]
\[ \Rightarrow n + 5 = 80 + 5 = 85\]
So, the three numbers are 81, 83 and 85.