Question

The sum of three consecutive even numbers is 276. Find the numbers.
A. 96, 92, 94
B. 91, 92, 94
C. 90, 94, 98
D. 90, 92, 94

Answer 1

Here we assume three consecutive even numbers as $(2n + 2),(2n + 4),(2n + 6)$. Add the three numbers and equate the sum to 276 to find the value of n. Substitute back the value of n in the numbers.

Complete step-by-step solution:
We have to take three consecutive natural numbers.
Let the three consecutive even numbers be $(2n + 2),(2n + 4),(2n + 6)$
We are given the sum of three consecutive even numbers is 249
Add the three numbers:
$\Rightarrow (2n + 2) + (2n + 4) + (2n + 6) = 276$
Add like values in LHS
$\Rightarrow 6n + 12 = 276$
Shift the constant values to RHS of the equation.
$\Rightarrow 6n = 276 - 12$
$\Rightarrow 6n = 264$
Divide both sides of the equation by 6
$\Rightarrow \dfrac{{6n}}{6} = \dfrac{{264}}{6}$
Cancel the same terms from numerator and denominator.
$\Rightarrow n = 44$
Now we substitute the value of n in each $(2n + 2),(2n + 4),(2n + 6)$ to obtain the three numbers.
Put$n = 44$in $(2n + 2)$
$\Rightarrow (2n + 2) = (2 \times 44 + 2)$
$\Rightarrow (2n + 2) = 88 + 2$
$\Rightarrow (2n + 2) = 90$
So, the first even number is 90.
Put$n = 44$in $(2n + 4)$
$\Rightarrow (2n + 4) = (2 \times 44 + 4)$
$\Rightarrow (2n + 4) = 88 + 4$
$\Rightarrow (2n + 4) = 92$
So, the second even number is 92.
Put$n = 44$in $(2n + 6)$
$\Rightarrow (2n + 6) = (2 \times 44 + 6)$
$\Rightarrow (2n + 6) = 88 + 6$
$\Rightarrow (2n + 6) = 94$
So, the third even number is 94.
Therefore, the three consecutive even numbers are 90, 92 and 94.

$\therefore$Option D is correct

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 $n,n + 2,n + 4$
Then the sum of three consecutive even numbers is 276
$\Rightarrow n + n + 2 + n + 4 = 276$
Add like values in LHS
$\Rightarrow 3n + 6 = 276$
Shift the constant values to RHS of the equation.
$\Rightarrow 3n = 276 - 6$
$\Rightarrow 3n = 270$
Divide both sides of the equation by 3
$\Rightarrow \dfrac{{3n}}{3} = \dfrac{{270}}{3}$
Cancel the same terms from numerator and denominator.
$\Rightarrow n = 90$
Substitute the value of n in $n + 2,n + 4$
$\Rightarrow n + 2 = 90 + 2 = 92$
$\Rightarrow n + 4 = 90 + 4 = 94$
So, the three numbers are 90, 92 and 94.
$\therefore$Option D is correct