Question

6 Find three consecutive numbers such that the sum of the first and the third is 36.​

Answer 1

Answer:

Let 2n = the first even number, where n is an integer.

Let 2n + 2 = the second even number, and ...

Let 2n + 4 = the third even number.

With the above three equalities in mind, the statement that, "The sum of three consecutive even numbers is 36" can be translated mathematically into the following equation to be solved for n:

2n + (2n + 2) + (2n + 4) = 36

2n + 2n + 2 + 2n + 4 = 36

Collecting like-terms on the left, we have:

6n + 6 = 36

6n + 6 - 6 = 36 - 6

6n + 0 = 30

6n = 30

(6n)/6 = 30/6

(6/6)n = 5

(1)n = 5

n = 5

Therefore, ...

2n = 2(5) = 10

2n + 2 = 10 + 2 = 12

2n + 4 = 10 + 4 = 14

CHECK:

2n + (2n + 2) + (2n + 4) = 36

2(5) + (2(5) + 2) + (2(5) + 4) = 36

10 + 12 + 14 = 36

36 = 36

Therefore, the three consecutive even numbers are: 10, 12, and 14.

Answer 2

Answer:

Step-by-step explanation:

let the 3 consecutive numbers be a-d,a.a+d

sum of 1st and 3rd number is

a-d-(a+d)=36

2(a)=36

therefore, a=18

now sum of all 3 numbers is

a-d+a+a+d=3a

54

sum of 3 numbers acc to A.P formula where a=18,n=3 Sn=54,

Sn=n/2[2a+(n-1)d]

54=3/2[2x18+2d]

54x2/3=36+2d

36=36+2d

therefore d=0

hence the 3 numbers are 18,18,18..

hope it was of some use