Question

Sum of 3 consecutive integers is 20 more than middle integer, find the smallest of three.

Answer 1

$let \: the \: consecutive \: intgers \: are \: x \: x + 1 \: and \: x + 2 \\ according \: to \: question \: sum = x + (x + 1) + (x + 2) = (x + 1) + 20 \\ 3x + 3 = x + 21 \\ 2x = 18 \\ x = 9 \\ therefore \: integers \: are \: 9 \: 10 \: and \: 11$

Answer 2

Given - Sum

Find - The smallest number

Solution - The smallest number among the three integers is 9

Let the three consecutive integers be x, x+1 and x+2.

Now, as per the given information, the mathematical equation that will form is -

x + x + 1 + x + 2 = 20 + x + 1

3x + 3 = x + 21

3x - x = 21 - 3

2x = 18

x= 18/2

x = 9

First integer = x

First integer = 9

Second integer = x + 1

Second integer = 9 + 1

Second integer = 10

Third integer = x + 2

Third integer = 9 + 2

Third integer = 11

Hence, the smallest of three integers is 9