Question

Find the sum of all natural numbers from \[100\] to \[300\].
(i) Which are divisible by \[5\].
(ii) Which are divisible by \[6\].
(iii) Which are divisible by \[5\] and \[6\].

Answer 1

Hint:In this question, we need to find the sum of all natural numbers from \[100\] to \[300\] under the given three conditions that which are divisible by \[5\] , which are divisible by \[6\] and which are divisible by \[5\] and \[6\]. This sum is based on the concept of A.P. Here A.P. stands for arithmetic progression. It is a sequence where the difference between the consecutive terms are the same. From the given series, we can find the first term \[(a)\] and the common difference \[(d)\] . Then , by using the formula of arithmetic progression, we can easily find \[n\]. Then by using the sum of n terms of A.P. we can find the required solution.

Formula used:
The Formula used to find the \[n^{th}\] terms in arithmetic progression is
\[a_{n} = \ a\ + \ \left( n\ -\ 1 \right) \times \ d\]
Where \[a\] is the first term, \[d\] is the common difference, \[n\] is the number of term and \[a_ n\] is the \[n^{th}\] term.
The formula for the summation of the terms is \[S_{n} = \dfrac{n}{2}\left( a + l \right)\ \] where \[l\] is the last term of the series.

Complete step by step answer:
First let us find the sum of all natural numbers from \[100\] to \[300\] which are divisible by \[5\].Now let us form the series from \[100\] to \[300\] which are divisible by \[5\],That is \[100,110,115\ldots 300\]. Here \[a\] is \[100\] , \[d\] is \[5\] and the last term \[l\] is \[300\].Now we need to find \[n\].
We can find \[n\] from \[a_{n} = \ a\ + \ \left( n\ -\ 1 \right) \times \ d\]
Where \[a_{n}\] is \[300\]
On substituting the known values,
We get,
\[\Rightarrow \ 300 = 100 + (n – 1) \times 5\]
On simplifying,
We get
\[300 = 100 + 5n – 5\]
On further simplifying,
We get,
\[300 = 5n + 95\]
On adding both sides by \[95\] ,
We get,
\[205 = 5n\]
On dividing both sides by \[5\] ,
We get
\[\Rightarrow \ n = \dfrac{205}{5}\]
On simplifying,
We get,
\[\Rightarrow \ n = 41\]

Now we can easily find the sum of the series which are divisible by \[5\] by using the summation formula.
That is \[S_{n} = \dfrac{n}{2}\left( a + l \right)\ \]
On substituting the values,
We get,
\[S_{41} = \dfrac{41}{2}(100 + 300)\]
On simplifying,
We get,
\[S_{41} = \dfrac{41}{2}\left( 400 \right)\]
On further simplifying,
We get,
\[S_{41} = 8200\]
Thus, the sum of all natural numbers from \[100\] to \[300\] which are divisible by \[5\] is \[8200\].

Next let us find the sum of all natural numbers from \[100\] to \[300\] which are divisible by \[6\].Now let us form the series from \[100\] to \[300\] which are divisible by \[6\].That is \[102,108,114\ldots 300\]. Here \[a\] is \[102\] , \[d\] is \[6\] and \[l\] is \[300\]. Now we need to find \[n\].
We can find \[n\] from \[a_{n} = \ a\ + \ \left( n\ -\ 1 \right) \times \ d\]
Where \[a_{n}\] is \[300\]
On substituting the known values,
We get,
\[300 = 102 + (n – 1) \times 6\]
On simplifying,
We get,
\[300 = 102 + 6n – 6\]
On further simplifying,
We get,
\[300 = 96 + 6n\]
On subtracting both sides by \[96\] ,
We get,
\[204 = 6n\ \]
On dividing both sides by \[6\] ,
We get,
\[\Rightarrow \ n = 34\]

Now we can easily find the sum of the series which are divisible by \[6\] by using the summation formula.
That is \[S_{n} = \dfrac{n}{2}(a + l)\]
On substituting the values,
We get,
\[S_{34} = \dfrac{34}{2}(102 + 300)\]
On simplifying,
We get,
\[S_{34} = \dfrac{34}{2} \times 402\]
On further simplifying,
We get,
\[S_{34} = 6834\]
Thus , the sum of all natural numbers from \[100\] to \[300\] which are divisible by \[6\] is \[6834\].Finally let us find the sum of all natural numbers from \[100\] to \[300\] which are divisible by \[5\] and \[6\] .

Now let us form the series between \[100\] to \[300\] which are divisible by \[5\] and \[6\] .
That is \[120,150,180,210,240,270,300\]. Here \[a\] is \[120\] , \[d\] is \[30\] and \[l\] is \[300\]. Here \[n\] can be found by counting the numbers that \[n\] is \[6\],Now we can easily find the sum of the series which are divisible by \[5\] and \[6\] by using the summation formula.
That is \[S_{n} = \dfrac{n}{2}(a + l)\]
Now on substituting the known values,
We get,
\[S_{7} = \dfrac{7}{2}(120 + 300)\]
On simplifying,
We get,
\[S_{7} = \dfrac{7}{2} \times 420\]
On further simplifying,
We get,
\[S_{7} = 1470\]
Thus the sum of all natural numbers from \[100\] to \[300\] which are divisible by \[5\] and \[6\] is \[1470\].

Hence, the final answer is:
The sum of all natural numbers from \[100\] to \[300\] which are divisible by \[5\] is \[8200\].
The sum of all natural numbers from \[100\] to \[300\] which are divisible by \[6\] is \[6834\].
The sum of all natural numbers from \[100\] to \[300\] which are divisible by \[5\] and \[6\] is \[1470\].

Note:In order to solve these types of questions, we should have a strong grip over arithmetic series. We should be very careful in choosing the correct formula because there is the chance of making mistakes in interchanging the formula of finding term and summation of term. If we try to solve this sum only with the formula \[S_{n} = \dfrac{n}{2}\left( a + l \right)\ \] where \[l\] is the last term of the series , without finding \[n\] , then our answer could not be found so we may get confused.