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Explanation :-

  • Let's assume the lowest integer be x.Then the next integer x+2 (since our integers need to be even) and the third integer x+4.

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It’s given–

  • The sum of three consecutive even numbers is 42.

[tex] \bf{\underline{\underline{\dag According\ to\ the\ question:-}}} [/tex]

[tex]\qquad[/tex][tex] \pink{\bf\longrightarrow {x+( x +2)+(x+4) = 42} }[/tex]

[tex]\qquad[/tex][tex] \sf \longrightarrow 3x +6 =42[/tex]

[tex]\qquad[/tex][tex] \sf \longrightarrow 3x = 42-6[/tex]

[tex]\qquad[/tex][tex] \sf \longrightarrow 3x =36[/tex]

[tex]\qquad[/tex][tex] \sf \longrightarrow x = \cancel {\dfrac{36}{3}}[/tex]

[tex]\qquad[/tex][tex]\pink{ \bf\longrightarrow x = 12}[/tex] ( which is our lowest integer.)

  • The smallest of these numbers is 12. And the next two integers we find with 12+2=14 and 12+4=16.

[tex]\qquad[/tex]____________________

Given :

  • Sum of three consecutive even numbers is 42. We are to find the smallest of the smallest consecutive even number.

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Solution :

Let us assume the three consecutive even numbers be "x" , "(x + 2)" and "(x + 4)" . Here x is the smallest number .

So,

According to the Question :

x + (x + 2) + (x + 4) = 42

⇢x + x + 2 + x + 4 = 42

⇢ x + x + x + 2 + 4 = 42

⇢ 3x + 6 ⠀⠀⠀⠀⠀ = 42

⇢ 3x ⠀⠀⠀⠀⠀⠀⠀⠀= 42 - 6

⇢ 3x ⠀⠀⠀⠀⠀⠀⠀⠀= 36

⇢ x ⠀⠀⠀⠀⠀⠀⠀⠀= 36/3

⇢ x ⠀⠀⠀⠀⠀⠀⠀⠀ = 12

Hence, 12 is the smallest consecutive even number.

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