Find the sum of all odd numbers between (i) 0 and 50 (ii) 100 and 200.

Solution:

(i) In this problem, we need to find the sum of all odd numbers lying between 0 and 50.

So, we know that the first odd number after 0 is 1 and the last odd number before 50 is 49.

Also, all these terms will form an A.P. with the common difference of 2.

So here,

First term (a) = 1

Last term (l) = 49

Common difference (d) = 2

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

$a_{e}=a+(n-1) d$

So, for the last term,

$49=1+(n-1) 2$

$49=1+2 n-2$

 

$49=2 n-1$

$49+1=2 n$

Further simplifying,

$50=2 n$

$n=\frac{50}{2}$

 

$n=25$

Now, using the formula for the sum of n terms,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

For $n=25$, we get,

On further simplification, we get,

$S_{n}=25(25)$

$=625$

Therefore, the sum of all the odd numbers lying between 0 and 50 is $S_{n}=625$.

(ii) In this problem, we need to find the sum of all odd numbers lying between 100 and 200.

So, we know that the first odd number after 0 is 101 and the last odd number before 200 is 199.

Also, all these terms will form an A.P. with the common difference of 2.

So here,

First term (a) = 101

Last term (l) = 199

Common difference (d) = 2

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

$a_{n}=a+(n-1) d$

So, for the last term,

$199=101+(n-1) 2$

$199=101+2 n-2$

 

$199=99+2 n$

$199-99=2 n$

Further simplifying,

$100=2 n$

$n=\frac{100}{2}$

$n=50$

Now, using the formula for the sum of n terms,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

For n = 50, we get,

$S_{n}=\frac{50}{2}[2(101)+(50-1) 2]$

$=25[202+(49) 2]$

$=25(202+98)$

$=25(300)$

 

$=7500$

Therefore, the sum of all the odd numbers lying between 100 and 200 is $S_{n}=7500$.