Question

Find three consecutive numbers such that the sum of the second and the third numbers exceeds the first by 14.​

Answer 1

Answer:

Let the first number be x

Other numbers will be (x + 1) and (x + 2)

Now, According to the question:

(x + 1) + (x + 2) = x + 14

x + 1 + x + 2 = x + 14

2x + 3 = x + 14

2x - x = 14 - 3

x = 11

Numbers will be 11

(x + 1) = 12

(x + 2) = 13

Hope this helps :)

Answer 2

The three consecutive numbers are "11", "12" and "13". Further explanation is given below.

Step-by-step explanation:

Let the numbers will  be:

x, (x+1) and (x+2)

Sum of 2nd and 3rd = (x+1)+(x+2)

Exceed first by 14 = x+14

Now,

⇒   x+1+x+2 = x+14

⇒   2x+3 = x+14

On subtracting "x" from both sides, we get

⇒   2x+3-x = x+14-x

⇒   x+3 = 14

⇒   x = 14-3

⇒   x = 11

First number = 11

Second number = x+1

                            = 11+1

                            = 12

Third number = x+2

                        = 11+2

                        = 13

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