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Problem: The sum of 4 consecutive even numbers is 36. What is the first number in this sequence?

Point to be noted: • In case of positive and negative concept the zero “0” is neither positive nor negative. • In case of even and odd concept zero”0” is even number. Because according to definition of even those numbers whose last number are 0 (or)2(or)4 (or)6(or)8 are called even numbers (or) simply we can also say that A number which is divisible by 2 and generates a remainder of 0 is called an even number.

• And also the difference between two consecutive even numbers will be 1 like if one is “n” then 2nd consecutive even will be “n+2”.

Solution: As given that we have 4 even numbers, so let us suppose, 1st even number=n 2nd even number=n+2 3rd even number=n+4 4th even number=n+6 Hence according to given condition

Sum of all the numbers = 36 1st +2nd +3rd +4th =36 n+n+2+n+4+n+6=36 add like terms with like terms we get,

4n+12=36 4n=36-12 4n=24 Divide both sides by 4 we get n=6

hence 1st even number =n=6 so 2nd even number=n+2=6+2=8 3rd even number=n+4=6+4=10 4th even number=n+6=6+6=12

Hence the sequence becomes 6,8,10,12 Where the 1st term of sequence is 6.

Verification: You can verify your answer simply by adding these terms

=6+8+10+12 =36 so it gives 36 and satisfy condition that’s given

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