Find the sum of all natural numbers between 1 and 100, which are divisible by 2 or 5.
Answer
330.9k+ views
Hint- Here, the formulas of ${n^{{\text{th}}}}$ term of an AP and sum of first $n$ terms of an AP will be used.
The natural numbers which are divisible by 2 are 2, 4, 6, 8, ……, 98, 100.
Clearly, the above series is an arithmetic progression with a common difference of 2.
The natural numbers which are divisible by 5 are 5, 10, 15, 20, ……, 95, 100.
Clearly, the above series is an arithmetic progression with a common difference of 5.
For an arithmetic progression consisting of first term as $a$, common difference as $d$ and total number of terms as $n$
Last term or ${n^{{\text{th}}}}$ term of the AP is given by ${a_n} = a + \left( {n - 1} \right)d{\text{ }} \to {\text{(1)}}$
Sum of first $n$ terms of the AP is given by ${{\text{S}}_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]{\text{ }} \to {\text{(2)}}$.
Now, for the arithmetic progression 2, 4, 6, 8, ……, 98, 100
$a = 2$, $d = 2$ and ${a_n} = 100$
Using equation (1), we can write
$ \Rightarrow 100 = 2 + 2\left( {n - 1} \right) \Rightarrow 100 = 2 + 2\left( {n - 1} \right) \Rightarrow 100 = 2 + 2n - 2 \Rightarrow 2n = 100 \Rightarrow n = 50$
So, the total number of natural numbers between 1 and 100 that are divisible by 2 are 50.
Now using equation (2), we have
${{\text{S}}_n} = \dfrac{{50}}{2}\left[ {2 + 100} \right] = \dfrac{{50 \times 102}}{2} = 2550$
So, the sum of all the natural numbers between 1 and 100 that are divisible by 2 is 2550.
Now, for the arithmetic progression 5, 10, 15, 20, ……, 95, 100
$a = 5$, $d = 5$ and ${a_n} = 100$
Using equation (1), we can write
$ \Rightarrow 100 = 2 + 2\left( {n - 1} \right) \Rightarrow 100 = 5 + 5\left( {n - 1} \right) \Rightarrow 100 = 5 + 5n - 5 \Rightarrow 5n = 100 \Rightarrow n = 20$
So, the total number of natural numbers between 1 and 100 that are divisible by 5 are 20.
Now using equation (2), we have
${{\text{S}}_n} = \dfrac{{20}}{2}\left[ {5 + 100} \right] = \dfrac{{20 \times 105}}{2} = 1050$
So, the sum of all the natural numbers between 1 and 100 that are divisible by 5 is 1050.
Also, natural numbers between 1 and 100 that are divisible by both 2 and 5 are 10, 20, 30, ……, 100
The above series also forms arithmetic progression with a common difference of 10.
Here, $a = 10$, $d = 10$ and ${a_n} = 100$
Using equation (1), we can write
$ \Rightarrow 100 = 10 + 10\left( {n - 1} \right) \Rightarrow 90 = 10n - 10 \Rightarrow 10n = 100 \Rightarrow n = 10$
So, the total number of natural numbers between 1 and 100 that are divisible by both 2 and 5 are 10.
Now using equation (2), we have
${{\text{S}}_n} = \dfrac{{10}}{2}\left[ {10 + 100} \right] = \dfrac{{10 \times 110}}{2} = 550$
So, the sum of all the natural numbers between 1 and 100 that are divisible by both 2 and 5 is 550.
Since, the sum of all the natural numbers between 1 and 100 that are divisible by 2 or 5 is equal to the sum of all the natural numbers between 1 and 100 that are divisible by 2 and those that are divisible by 5 minus the number of natural numbers between 1 and 100 that are divisible by both 2 and 5.
i.e., ${\text{Required Sum}} = \left( {2550 + 1050} \right) - 550 = 3050$.
Therefore, the sum of all natural numbers between 1 and 100, which are divisible by 2 or 5 is 3050.
Note- In this particular problem, the sum of numbers between 1 and 100 that are divisible by 2 and those that are divisible by 5 will include the numbers that are divisible by both 2 and 5 twice. So in order to find the accurate result we will once subtract the sum of all the numbers between 1 and 100 that are divisible by both 2 and 5.
The natural numbers which are divisible by 2 are 2, 4, 6, 8, ……, 98, 100.
Clearly, the above series is an arithmetic progression with a common difference of 2.
The natural numbers which are divisible by 5 are 5, 10, 15, 20, ……, 95, 100.
Clearly, the above series is an arithmetic progression with a common difference of 5.
For an arithmetic progression consisting of first term as $a$, common difference as $d$ and total number of terms as $n$
Last term or ${n^{{\text{th}}}}$ term of the AP is given by ${a_n} = a + \left( {n - 1} \right)d{\text{ }} \to {\text{(1)}}$
Sum of first $n$ terms of the AP is given by ${{\text{S}}_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]{\text{ }} \to {\text{(2)}}$.
Now, for the arithmetic progression 2, 4, 6, 8, ……, 98, 100
$a = 2$, $d = 2$ and ${a_n} = 100$
Using equation (1), we can write
$ \Rightarrow 100 = 2 + 2\left( {n - 1} \right) \Rightarrow 100 = 2 + 2\left( {n - 1} \right) \Rightarrow 100 = 2 + 2n - 2 \Rightarrow 2n = 100 \Rightarrow n = 50$
So, the total number of natural numbers between 1 and 100 that are divisible by 2 are 50.
Now using equation (2), we have
${{\text{S}}_n} = \dfrac{{50}}{2}\left[ {2 + 100} \right] = \dfrac{{50 \times 102}}{2} = 2550$
So, the sum of all the natural numbers between 1 and 100 that are divisible by 2 is 2550.
Now, for the arithmetic progression 5, 10, 15, 20, ……, 95, 100
$a = 5$, $d = 5$ and ${a_n} = 100$
Using equation (1), we can write
$ \Rightarrow 100 = 2 + 2\left( {n - 1} \right) \Rightarrow 100 = 5 + 5\left( {n - 1} \right) \Rightarrow 100 = 5 + 5n - 5 \Rightarrow 5n = 100 \Rightarrow n = 20$
So, the total number of natural numbers between 1 and 100 that are divisible by 5 are 20.
Now using equation (2), we have
${{\text{S}}_n} = \dfrac{{20}}{2}\left[ {5 + 100} \right] = \dfrac{{20 \times 105}}{2} = 1050$
So, the sum of all the natural numbers between 1 and 100 that are divisible by 5 is 1050.
Also, natural numbers between 1 and 100 that are divisible by both 2 and 5 are 10, 20, 30, ……, 100
The above series also forms arithmetic progression with a common difference of 10.
Here, $a = 10$, $d = 10$ and ${a_n} = 100$
Using equation (1), we can write
$ \Rightarrow 100 = 10 + 10\left( {n - 1} \right) \Rightarrow 90 = 10n - 10 \Rightarrow 10n = 100 \Rightarrow n = 10$
So, the total number of natural numbers between 1 and 100 that are divisible by both 2 and 5 are 10.
Now using equation (2), we have
${{\text{S}}_n} = \dfrac{{10}}{2}\left[ {10 + 100} \right] = \dfrac{{10 \times 110}}{2} = 550$
So, the sum of all the natural numbers between 1 and 100 that are divisible by both 2 and 5 is 550.
Since, the sum of all the natural numbers between 1 and 100 that are divisible by 2 or 5 is equal to the sum of all the natural numbers between 1 and 100 that are divisible by 2 and those that are divisible by 5 minus the number of natural numbers between 1 and 100 that are divisible by both 2 and 5.
i.e., ${\text{Required Sum}} = \left( {2550 + 1050} \right) - 550 = 3050$.
Therefore, the sum of all natural numbers between 1 and 100, which are divisible by 2 or 5 is 3050.
Note- In this particular problem, the sum of numbers between 1 and 100 that are divisible by 2 and those that are divisible by 5 will include the numbers that are divisible by both 2 and 5 twice. So in order to find the accurate result we will once subtract the sum of all the numbers between 1 and 100 that are divisible by both 2 and 5.
Last updated date: 31st May 2023
•
Total views: 330.9k
•
Views today: 7.87k
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
