Question

The sum of three consecutive odd numbers is 903. What will be $ \dfrac{{11}}{7} $ of the middle number?

Answer 1

Hint: The general form of an odd number is $ 2n + 1 $ where n belongs to whole numbers. When $ 2n + 1 $ is an odd number, then the next two odd numbers from this will be $ 2n + 1 + 2 = 2n + 3 $ , $ 2n + 3 + 2 = 2n + 5 $ . Add these three numbers and equate it to 903, you will find the n values. By using the value of n find the middle number.

Complete step-by-step answer:
We are given that the sum of three consecutive odd numbers is 903.
We have to find the $ \dfrac{{11}}{7} $ of the middle odd number.
Let the first odd number be $ 2n + 1,n \in W $ ‘W’ represent the whole numbers.
Then the second odd number will be $ 2n + 1 + 2 = 2n + 3 $
And the third odd number will be $ 2n + 3 + 2 = 2n + 5 $
So, now $ 2n + 1,2n + 3,2n + 5 $ are three consecutive odd numbers.
And the sum of these consecutive odd numbers is 903.
 $
  \left( {2n + 1} \right) + \left( {2n + 3} \right) + \left( {2n + 5} \right) = 903 \\
  2n + 2n + 2n + 1 + 3 + 5 = 903 \\
  6n + 9 = 903 \\
  6n = 903 - 9 \\
  6n = 894 \\
  n = \dfrac{{894}}{6} \\
  n = 149 \\
  $
The value of n is 149
The 1st odd number is $ 2n + 1 = 2\left( {149} \right) + 1 = 298 + 1 = 299 $
The 2nd (middle) odd number is $ 2n + 3 = 2\left( {149} \right) + 3 = 298 + 3 = 301 $
The 3rd odd number is $ 2n + 5 = 2\left( {149} \right) + 5 = 303 $
The middle number is 301.
 $ \dfrac{{11}}{7} $ of the middle number (301) is
 $
   = \dfrac{{11}}{7} \times 301 \\
   = 11 \times 43 \\
   = 473 \\
  $
Therefore, $ \dfrac{{11}}{7} $ of the middle number of the three consecutive odd numbers is 473.

Note: Numbers which follow each other in order, without any breaks, from smallest to largest are known as consecutive numbers. Odd consecutive numbers and even consecutive numbers have a difference of 2 between any two of those consecutive numbers.