Login/Sign Up
Home » Find the(i) Sum of those integers between 1 and 500 which are multiples of 2 as well as of 5. (ii) Sum of those integers from 1 to 500 which are multiples of 2 as well as of 5 .
Find the
(i) Sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.
(ii) Sum of those integers from 1 to 500 which are multiples of 2 as well as of 5 .
Arithmetic Progression | CBSE | Class 10 | Exercise 5.4 | Maths | NCERT Exemplar
Find the
(i) Sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.
(ii) Sum of those integers from 1 to 500 which are multiples of 2 as well as of 5 .

Solution:

(i) It is known to us that,

LCM of (2, 5) = 10 for multiples of 2 and 5.

Between 1 and 500, multiples of 2 and 5 = 10, 20, 30,…, 490.

As a result,

We can say that 10, 20, 30,…, 490 is an AP with a common difference of d = 10.

The first term, a = 10

Let’s assume the number of terms in this AP = n

Using the nth term formula,

{{a}_{n}}~=\text{ }a+\left( n-1 \right)d

490\text{ }=\text{ }10\text{ }+\text{ }\left( n-1 \right)10

480=\left( n-1 \right)10

n-1\text{ }=\text{ }48

n\text{ }=\text{ }49

The sum of an AP,

{{S}_{n}}~=\text{ }\left( n/2 \right)\text{ }[a\text{ }+\text{ }{{a}_{n}}], [here it is given that an is the last term]

=\text{ }\left( 49/2 \right)\text{ }\times \text{ }\left[ 10\text{ }+\text{ }490 \right]

=\text{ }\left( 49/2 \right)\text{ }\times \text{ }\left[ 500 \right]

=\text{ }49\text{ }\times \text{ }250

=\text{ }12250

As a result, 12250 is the sum of integers between 1 and 500 which are multiples of 2 as well as of 5.

(ii) It is known us to that,

LCM of (2, 5) = 10 for multiples of 2 and 5.

Between 1 and 500 multiples of 2 and 5 = 10, 20, 30…, 500.

As a result,

We can say that 10, 20, 30…, 500 is an AP with a common difference of d = 10

The first term, a = 10

Let’s assume the number of terms in this AP = n

Using the nth term formula,

{{a}_{n}}~=a+\left( n-1 \right)d

500=10+\left( n-1 \right)10

490=\left( n-1 \right)10

n-1=49

n\text{ }=\text{ }50

The sum of an AP,

{{S}_{n}}~=\text{ }\left( n/2 \right)\text{ }[\text{ }a\text{ }+\text{ }{{a}_{n}}], [here it is given that an is the last term]

=\text{ }\left( 50/2 \right)\text{ }\times \left[ 10+500 \right]

=\text{ }25\times \text{ }\left[ 10\text{ }+\text{ }500 \right]

=\text{ }25\left( 510 \right)

=\text{ }12750

As a result, 12750 is the sum of integers from 1 to 500 which are multiples of 2 as well as of 5.

i

More Material