Question

Find a set of 99 odd numbers whose sum is \[2010\].

Answer 1

Hint: The given condition is in arithmetic progression, use formula, to find the set.
We are going to assume that the first odd number of the set is $x$, and then we are going to write a finite arithmetic expression until 99 terms. Then use to find the value of x, such that we can find the whole set whose sum equals 2010.

Complete step by step answer:
We first have to assume that the first odd number of the set is $x$.
Then we form an arithmetic progression form until the ${99^{th}}$ term. Which is
$x + (x + 2) + (x + 4).............n$
Since we know that this follows arithmetic progress.
So,
$
  a = x \\
  d = 2 \\
  n = 99 \\
 $
Also, we are given that
${S_n} = 2010$
We, can use this find the value of $x$,
The formula for the sum of the arithmetic progression is
${S_n} = \dfrac{n}{2}(2a + (n - 1)d) - - (i)$
Now we substitute the value of the obtained values in equation (i), we get
$
  {S_{99}} = \dfrac{{99}}{2}(2\left( x \right) + (99 - 1)2)\,\,(\because n = 99) \\
  2010 = \dfrac{{99}}{2}(2\left( x \right) + (99 - 1)2) \\
 $
We further solve the above equation
$
   \Rightarrow 2010 = \dfrac{{99}}{2}\left( {2(x + 98)} \right) \\
   \Rightarrow 2010 = 99(x + 98) \\
   \Rightarrow 2010 = 99x + 9702 \\
   \Rightarrow 99x = 2010 - 9702 \\
   \Rightarrow 99x = - 7692 \\
 $
By further solving the above obtained equation, we get
$
  x = \dfrac{{ - 7692}}{{99}} \\
  x = - 77.69 \\
 $
From the last line we can see that the value obtained is negative $\left( {x < 0} \right)$.
Hence we can say the sum of the set of \[99\] odd numbers whose sum is equal to \[2010\] is not possible so there are no such numbers.

Note: The difference between any two consecutive odd numbers and the consecutive even numbers is 2. In this question we assumed a set of 99 numbers where the difference between the consecutive numbers are 2 since the given set of numbers are odd numbers.