Question

Find the three consecutive numbers whose sum is 108.
$$\begin{gathered} \text{A}\text{. 35,46,37} \\ \text{B}\text{. 5,36,37} \\ \text{C}\text{. 35,36,37} \\ \text{D}\text{. None of these} \end{gathered}$$

Answer 1

Hint- The series of numbers that are more than 1 from its preceding number id known as
consecutive numbers. The consecutive number series can be even as well as odd depending on the first
term of the series and the difference between the numbers. If the difference between the two
consecutive numbers is 2 and the starting term is an odd integer then, it is known as the series of the
odd consecutive numbers, whereas if the difference between the two consecutive numbers is 2 and the starting term is an even integer, then it is known to be an even consecutive number series. For example,
1,2,3,4,… is a series of general, consecutive numbers.
2,4,6,8,… is a series of even consecutive numbers.
1,3,5,7,… is a series of odd consecutive numbers.
Here, the question is asking for the series of consecutive integers whose sum is 108. So, consider the first
term to be x, and then accordingly, the second and the third term will be (x+1) and (x+2), respectively.
Complete step by step solution:
As $x,(x+1),(x+2)$ are consecutive numbers so, the summation should be equal to 108.
$$\begin{gathered} x+x+1+x+2=108 \\ 3x+3=108 \\ 3(x+1)=108 \\ x+1={\displaystyle \frac{108}{3}} \\ x+1=36 \\ x=35 \\ x+1=35+1=36 \\ x+2=36+1=37 \end{gathered}$$
Hence, 35,36,37 are the three consecutive numbers whose sum is 108.
Option D. is correct.

Note: It is interesting to note here that for a series of even or odd consecutive numbers, all the terms
present in the series are even and odd, respectively. Alternatively, if we see the option, only one option
consists of a series of consecutive numbers.