find three consecutive even consecutive numbers such that the sum of first and last numbers exceeds the second number by 10
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let the three consecutive even number be :x , x+2 , x+4.
the first number is "x"
the second number is "x+2"
the third number is "x+2+2"that is "x+4"
the sum of first and the last is 10 more than second.
=>(x)+(x+4)=(x+2)+10
=>x=8;
the second number is "x+2" is "8+2"=>10
the third number is "x+4"is "8+4"=>12
hence the three numbers are 8,10,12
the first number is "x"
the second number is "x+2"
the third number is "x+2+2"that is "x+4"
the sum of first and the last is 10 more than second.
=>(x)+(x+4)=(x+2)+10
=>x=8;
the second number is "x+2" is "8+2"=>10
the third number is "x+4"is "8+4"=>12
hence the three numbers are 8,10,12
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