Question

What is the sum of all odd numbers between \[500\] and \[600\] ?
\[\left( 1 \right)\] \[26950\]
\[\left( 2 \right)\] \[27500\]
\[\left( 3 \right)\] \[27950\]
\[\left( 4 \right)\] \[26500\]

Answer 1

Hint: We have to find the sum of all the odd numbers between \[500\] and \[600\] . We solve this question using the concept of sum of \[n\] number of terms of an arithmetic progression (A.P.) . We will also use the concept of finding the number of terms in the given series using the concept of the \[{n^{th}}\] term of an A.P. . Using the relation between the first term and last term of an A.P. , we will find the value of \[n\] (number of terms in the series) . Also from the data we will find the value of the first term and the common difference . Putting all these values in the formula of sum of \[n\] terms of A.P. we will get the value of sum of all odd numbers between \[500\] and \[600\] .

Complete step-by-step solution:
For the terms of a given series to be in A.P. the common difference between the terms of the series should be the same for all the two consecutive terms of the series . The difference of the second term and the first term of the given series should be the same as that of the difference of the third term and the second term of the given series .
Given :
Form the given data , we can compute the series of all the odd numbers between \[500\] and \[600\] to be as given below :
\[501{\text{ }},{\text{ }}503{\text{ }},{\text{ }}505{\text{ }},{\text{ }} \ldots \ldots \ldots \ldots \ldots \ldots \ldots {\text{ }},{\text{ }}599\]
Let the first term of the series be \[{a_1}\] , common difference of the series be \[d\] and the \[{n^{th}}\] term of the series be \[{a_n}\] (where n is the number of terms of the series) , \[{s_n}\] be the sum of the \[n\] number of terms of the series .
From the series , we get that
\[{a_1} = 501\]
\[{a_2} = 503\]
\[{a_3} = 505\]
\[{a_n} = 599\]
We know that the formula for the common difference is given as :
\[d = {a_2} - {a_1}\]
So , using the formula of the common difference , we get the value of common difference as:
\[d = 503 - 501\]
\[d = 2\]
We also , know that the formula for \[{n^{th}}\] term of A.P. is given as :
\[{a_n} = a + (n - 1)d\]
Putting the values in the formula , we get the value of \[n\] as
\[599 = 501\left( {n - 1} \right) \times 2\]
On solving ,
\[\Rightarrow (n - 1) \times 2 = 98\]
\[\Rightarrow (n - 1) = 49\]
\[\Rightarrow n = 50\]
Now , we know that the formula for sum of \[n\] terms of A.P. is given as :
\[{S_n} = \dfrac{n}{2}[2a + \left( {n - 1} \right)d]\]
Putting the values in the formula , we get the value of sum of all the odd numbers between \[500\] and \[600\] as :
\[{s_n} = \dfrac{{50}}{2}[2 \times 501 + \left( {50 - 1} \right)2]\]
\[\Rightarrow {S_n} = 50\left[ {501 + 49} \right]\]
On solving , we get
\[{S_n} = 27500\]
Thus , the sum of all the odd numbers between \[500\] and \[600\] is \[27500\] .
Hence , the correct option is \[(2)\] .

Note: While writing the series for the given condition , we would carefully note the statement given in the question . As stated in the question all the odd numbers between \[500\] and \[600\] , so we will be taking all the odd numbers between \[500\] and \[600\] but excluding the two numbers . And further solve the question using the series.