Question

Find the sum of all natural numbers between 200 and 300 which are exactly divisible by 6.

Answer 1

Hint: First, we will write the series between 200 and 300. Then using the divisibility rule i.e., we have to take even numbers and addition of the digits of that number should be divisible by 3. Then that number is exactly divisible by 6. Using this we will get a series which will be in front of Arithmetic progression. So, to find the total number of terms in series, we will use the formula ${{T}_{n}}=a+\left( n-1 \right)d$ . Thus, after getting the number of terms n, we will use the summation formula ${{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$ and will get the answer.

Complete step-by-step solution -
Here, we will see that the series will be $201,202,203,204,......299$ as it is told between 200 and 300 so, we will not include both numbers in series.
Now, here we will use the divisibility rule of 6 i.e. we have to take even numbers and addition of the digits of that number should be divisible by 3. Then that number is exactly divisible by 6.
So, here we will take 202 which is even but adding the digits is equal to 4 which cannot be divided by 3. So, this number will not be divisible by 6.
Now, taking 204, whose summation is equal to 6 which is divisible by 3. So, this is the first term. Similarly, finding this way we get the series as
$204,210,216,....294$ ………………………………(1)
We can see that this is in the Arithmetic progression series whose first term $a=204$ , common difference $d=210-204=6$ and last term ${{T}_{n}}=294$ . So, we will use the formula for finding the number of terms present in series using the formula ${{T}_{n}}=a+\left( n-1 \right)d$ . On substituting the values, we get as
${{T}_{n}}=a+\left( n-1 \right)d$
$294=204+\left( n-1 \right)6$
On further solving, we get
$294-204=6n-6$
$90+6=6n$
On dividing the equation by 6 we get,
$\dfrac{96}{6}=n=16$
Thus, there are a total 16 terms in the series $204,210,216,....294$ which are exactly divisible by 6.
Now, to find summation of all the terms, we will use the formula ${{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$ , where n is total number of terms i.e. 16, a is first term of series i.e. 204 and l is last term of series which is 294.
On substituting the values in the equation, we get as
${{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$
${{S}_{n}}=\dfrac{16}{2}\left( 204+294 \right)$
On solving, we get
${{S}_{n}}=8\left( 498 \right)=3984$
Thus, the sum of all-natural numbers between 200 and 300 which are exactly divisible by 6 is 3984.

Note: Do not include 300 numbers in the series. Sometimes students make mistakes here, instead of not including they include 300 which is exactly divisible by 6. By doing this the whole answer will be changed and the series will be like this $204,210,216,....294,300$ . So, number of terms will be equal to 17 and summation will be ${{S}_{n}}=\dfrac{17}{2}\left( 204+300 \right)=\dfrac{8568}{2}=4284$ . So, do not make this mistake and it will result in a wrong answer.