Question

Find the sum of the integers between $100$ and $200$ that are divisible by $6$ ?

Answer 1

In this problem we have to find the sum of the integers between $100$ and $200$ that are divisible by $6$. Then we will find the first divisible of $6$in between the $100$ and $200$. That is $102$. Then we will find the last divisible of $6$in between the $100$ and $200$. That is $198$. The number between $100$ and $200$ can be obtained by adding $6$to the previous divisible number, we can write all the numbers divisible by $6$in arithmetic progression. Then we will use the ${{n}^{th}}$term formula. That is ${{t}_{n}}=a+(n-1)d$. we will take $a$term as first divisible of $6$,$d$is the difference in between the $100$ and $200$and ${{t}_{n}}$ is the last divisible of $6$. Now we will find $n$value. then we will use another formula that is $sum=\dfrac{n}{2}\left[ a+{{a}_{n}} \right]$ then we will get the final result.

Formula used:
1.${{n}^{th}}$term formula that is ${{t}_{n}}=a+(n-1)d$.
2. $sum=\dfrac{n}{2}\left[ a+{{a}_{n}} \right]$.

Complete step by step solution:
Given that, the sum of the integers between $100$ and $200$ that are divisible by $6$.
Now we will take the first and last divisible of the $6$in between the number of $100$ and $200$.
The first one is $102$ and last one is $198$.
Now we will convert the arithmetic progression method, then
$102,102+6,102+6+6,.....$.
Now we will use the ${{n}^{th}}$term formula that is
${{t}_{n}}=a+(n-1)d$.
Now we take
$a$is the first divisible of $6$ that is $102$
$d$is the $6$.
${{t}_{n}}$is the last divisible of $6$that is $198$.
Now we will simplify the ${{n}^{th}}$ terms
$198=102+(n-1)6$
Now we will simplify the above equation that is
$\begin{aligned} & \Rightarrow \left( n-1 \right)=198-102 \\\ & \Rightarrow \left( n-1 \right)=96 \\\ \end{aligned}$
Now we will find $n$ value that is
$n=17$.
Now we will use another formula that is
$sum=\dfrac{n}{2}\left[ a+{{a}_{n}} \right]$.
Now we will simplify the above formula that is
$\begin{aligned} & \Rightarrow sum=\dfrac{17}{2}\left[ 102+198 \right] \\\ & \Rightarrow sum=\dfrac{17}{2}\times 300 \\\ & \Rightarrow sum=150\times 17 \\\ & \Rightarrow sum=2550 \\\ \end{aligned}$

Hence the required result is $2550$.

Note:
In this problem they have asked to find the sum of the numbers divided by $6$. So, we have developed an arithmetic progression with a common difference $6$. If they have asked to find the sum of numbers divisible by $n$ then we need to develop an arithmetic progression with common difference $n$.