How do you find four consecutive integers whose sum is 114?

Hint: In this question, we have to find the four consecutive numbers whose sum is equal to 114. Consecutive numbers are those numbers that are next to each other, that is, one number is one more than its previous one, or we can also suppose one number to be smaller than the next one. By letting one number be “x”, the other number can be supposed to be “x+1” or “x-1” and so on. Then the sum of these two numbers is given as 114, so we can form an equation and solve the obtained equation using the appropriate method.

Complete step-by-step solution:
Let one number be “x” then the other numbers are x+1, x+2 and x+3.
We are given that –
$x + x + 1 + x + 2 + x + 3 = 114$
We will group the like terms and apply the arithmetic operations –
$
   \Rightarrow x + x + x + x + 1 + 2 + 3 = 114 \\
   \Rightarrow 4x + 6 = 114 \\
 $
Now, we will take 6 and 4 to the right-hand side –
$
   \Rightarrow 4x = 114 - 6 \\
   \Rightarrow 4x = 108 \\
   \Rightarrow x = \dfrac{{108}}{4} = 27 \\
 $
So the first integer is $27$ , the second integer is $27 + 1 = 28$ , the third integer is $27 + 2 = 29$ and the fourth integer is $27 + 3 = 30$ .
Hence, the four consecutive integers whose sum is 114 are 27, 28, 29 and 30.

Note: The equation obtained is an algebraic equation as it contains a combination of alphabets and numerical values. Now, for finding “n” number of unknown quantities, we require “n” number of equations. In this question, we have exactly one equation and one unknown quantity so its value is solved easily by rewriting the equation in such a way that the terms containing “x” remain on one side and the constant values on the other side.