Answer
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Hint: 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
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
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