
How do you add all the odd numbers between $1$ and $99$ inclusive ?
Answer
541.2k+ views
Hint: In the given question, we are required to find the sum of all the odd numbers between $1$ and $99$ including both $1$ and $99$. The question can be solved using the concepts of arithmetic progression. The problem requires us to have knowledge about finding the sum of n terms of an arithmetic series.
Complete step by step answer:
So, we have to find the sum $1 + 3 + 5 + 7 + ...... + 97 + 99$.
We can see that the given numbers are in arithmetic progression.
Now, we have to find the common difference of the arithmetic progression.
Let the common difference be d.
Then, $d = \left( {3 - 2} \right) = \left( {5 - 3} \right) = \left( {99 - 97} \right) = 2$
Now, we know that the nth term of the series can be found using the formula ${a_n} = a + \left( {n - 1} \right)d$.
So, $99 = 1 + \left( {n - 1} \right)\left( 2 \right)$
Subtracting $1$ from both sides of the equation, we get,
$ \Rightarrow 2\left( {n - 1} \right) = 99 - 1$
Simplifying the calculation further, we get,
$ \Rightarrow 2\left( {n - 1} \right) = 98$
$ \Rightarrow \left( {n - 1} \right) = 49$
Simplifying further, we get,
$ \Rightarrow n = 50$
So, $99$ is the ${50^{th}}$ term in the arithmetic progression.
Now, we know the number of terms in the series as well that is required to find the sum of the progression.
The sum of n terms of an arithmetic progression is $\dfrac{n}{2}\left( {2\left( a \right) + \left( {n - 1} \right)\left( d \right)} \right)$.
Therefore, $1 + 3 + 5 + 7 + ...... + 97 + 99 = \dfrac{{50}}{2}\left( {2\left( 1 \right) + \left( {50 - 1} \right)\left( 2 \right)} \right)$
$ \Rightarrow 1 + 3 + 5 + 7 + ...... + 97 + 99 = 25\left( {2 + 98} \right)$
Simplifying further, we get,
$ \Rightarrow 1 + 3 + 5 + 7 + ...... + 97 + 99 = 25\left( {100} \right)$
$ \Rightarrow 1 + 3 + 5 + 7 + ...... + 97 + 99 = 2500$
So, the sum of all the odd numbers between $1$ and $99$ including both $1$ and $99$ is $2500$.
Note: There are various methods to find the summation required in the given problem. The method discussed in the solution is the standard method to solve such questions and should be kept in mind as it can be used to solve other questions of the same type.
Complete step by step answer:
So, we have to find the sum $1 + 3 + 5 + 7 + ...... + 97 + 99$.
We can see that the given numbers are in arithmetic progression.
Now, we have to find the common difference of the arithmetic progression.
Let the common difference be d.
Then, $d = \left( {3 - 2} \right) = \left( {5 - 3} \right) = \left( {99 - 97} \right) = 2$
Now, we know that the nth term of the series can be found using the formula ${a_n} = a + \left( {n - 1} \right)d$.
So, $99 = 1 + \left( {n - 1} \right)\left( 2 \right)$
Subtracting $1$ from both sides of the equation, we get,
$ \Rightarrow 2\left( {n - 1} \right) = 99 - 1$
Simplifying the calculation further, we get,
$ \Rightarrow 2\left( {n - 1} \right) = 98$
$ \Rightarrow \left( {n - 1} \right) = 49$
Simplifying further, we get,
$ \Rightarrow n = 50$
So, $99$ is the ${50^{th}}$ term in the arithmetic progression.
Now, we know the number of terms in the series as well that is required to find the sum of the progression.
The sum of n terms of an arithmetic progression is $\dfrac{n}{2}\left( {2\left( a \right) + \left( {n - 1} \right)\left( d \right)} \right)$.
Therefore, $1 + 3 + 5 + 7 + ...... + 97 + 99 = \dfrac{{50}}{2}\left( {2\left( 1 \right) + \left( {50 - 1} \right)\left( 2 \right)} \right)$
$ \Rightarrow 1 + 3 + 5 + 7 + ...... + 97 + 99 = 25\left( {2 + 98} \right)$
Simplifying further, we get,
$ \Rightarrow 1 + 3 + 5 + 7 + ...... + 97 + 99 = 25\left( {100} \right)$
$ \Rightarrow 1 + 3 + 5 + 7 + ...... + 97 + 99 = 2500$
So, the sum of all the odd numbers between $1$ and $99$ including both $1$ and $99$ is $2500$.
Note: There are various methods to find the summation required in the given problem. The method discussed in the solution is the standard method to solve such questions and should be kept in mind as it can be used to solve other questions of the same type.
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