Question
Find the sum of all odd numbers between:
$100$ and $200$.
$100$ and $200$.
✓
Answer
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 from 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,
$199 = 101 + (n - 1)2$
$199 = 101 + 2n - 2$
$199 = 99 + 2n$
$199 - 99 = 2n$
Further simplifying,
$100 = 2n$
$\text{n}=\frac{100}{2}$
$n = 50$
Now, using the formula for the sum of n terms.
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{S}_\text{n}=\frac{50}{2}[2(101)+(50-1)2]$
$= 25[202 + (49)2]$
$= 25(202 + 98)$
$= 25(300)$
$= 7500$
therefore, the sum of all the odd numbers lying between $100$ and $200$ is $S_n = 7500.$
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 from 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,
$199 = 101 + (n - 1)2$
$199 = 101 + 2n - 2$
$199 = 99 + 2n$
$199 - 99 = 2n$
Further simplifying,
$100 = 2n$
$\text{n}=\frac{100}{2}$
$n = 50$
Now, using the formula for the sum of n terms.
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{S}_\text{n}=\frac{50}{2}[2(101)+(50-1)2]$
$= 25[202 + (49)2]$
$= 25(202 + 98)$
$= 25(300)$
$= 7500$
therefore, the sum of all the odd numbers lying between $100$ and $200$ is $S_n = 7500.$
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