Question

26. When added, four consecutive numbers have a sum of 18. What are the numbers?​

Answer 1

Answer:

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It isn't possible.

The reason that it is impossible because in order to add four consecutive odd integers to the sum of 1818, then one must also be able to add two odd integers (consecutive or not) to the sum of 9.9. Adding any two odd integers integers together will always result in an even sum, so a sum of 99 would be impossible.

It isn't even possible to add even twoconsecutive odd integers to equal 18.18. Here, let me show you.

Assume that nn is even. So two consecutive integers would then be (n−1) and (n+1).(n−1) and (n+1).

(n−1)+(n+1)=18(n−1)+(n+1)=18

⟹2n=18⟹2n=18

⟹n=9⟹n=9

So nn cannot be even, and two consecutive odd integers alone cannot sum to 18.18. This can be shown with the following.

7+9=167+9=16

9+11=209+11=20

But we are adding four consecutive odd integers. So:

(n−3)+(n−1)+(n+1)+(n+3)=18(n−3)+(n−1)+(n+1)+(n+3)=18

⟹4n=18⟹4n=18

Now obviously 1818 isn't divisible by 44 but it implies that (n−1)+(n+1)=9(n−1)+(n+1)=9 and (n−3)+(n+3)=9(n−3)+(n+3)=9 which is not possible for two odd integer to sum to. Just the same, let's add four consecutive odd integers.

1+3+5+7=161+3+5+7=16

3+5+7+9=243+5+7+9=24

Thus, what you ask for is impossible.

Q.E.D.

Answer 2

Hello Alish.. jee

Let the first number be 'a'

So, second number be 'a+1'

third number be 'a+2'

fourth number be 'a+3'

Then, sum of all = '4a+6'

Since, according to the question, sum is 18.

So, (4a+6) is equal to 18.

》4a+6 = 18

》4a = 12

》a = 3

So, Numbers = 3,4,5,6

Thanku...