Question

The sum of 5 consecutive odd numbers is 135. Find the numbers.

Answer 1

Hint: We must assume the numbers in such a way that most of the terms get cancelled. For this, we can assume the 5 numbers to be $\left( x-4 \right),\left( x-2 \right),x,\left( x+2 \right)\text{ and }\left( x+4 \right)$. Then by forming the equation of their sum being equal to 135, we can find the value of $x$, and thus each of the numbers separately.

Complete step-by-step solution:
We are given that the sum of 5 consecutive odd numbers is 135, and we have to find all these numbers.
We know that all consecutive odd numbers occur at an interval of 2.
So, let us assume that the numbers are $\left( x-4 \right),\left( x-2 \right),x,\left( x+2 \right)\text{ and }\left( x+4 \right)$.
According to the information given in the question, we can write
$\left( x-4 \right)+\left( x-2 \right)+x+\left( x+2 \right)+\left( x+4 \right)=135$
On opening the brackets, we can write
$x-4+x-2+x+x+2+x+4=135$
Hence, we simplify the above equation to get
$5x=135$
Now, we can divide both sides of this equation by 5. Thus, we get
$\dfrac{5x}{5}=\dfrac{135}{5}$
On simplification, we get
$x=27$
Thus, the value of number $\left( x-4 \right)$ is (27 – 4) = 23.
We know that the value of number $\left( x-2 \right)$ is (27 – 2) = 25.
The number $x$ is 27.
We can also say that the number $\left( x+2 \right)$ is (27 + 2) = 29.
Also, we see that the value of number $\left( x+4 \right)$ is (27 + 4) = 31.
Hence, the 5 numbers are 23, 25, 27, 29 and 31.

Note: We can also assume the numbers to be $x,\left( x+2 \right),\left( x+4 \right),\left( x+6\right)\text{ and }\left( x+8 \right)$. The rest of the process will be similar. But we must note that this method is a slightly longer approach that the one solved above. We must understand that if there would have been even numbers, like 4 or 6 instead of 5, we must have assumed the numbers to be $...,\left( x-3 \right),\left( x-1 \right),\left( x+1 \right),\left( x+3 \right),...$Using such assumptions, we can solve similar problems very easily.