Question

find the sum of all natural numbers between 50 and 150 which are multiples of 6​

Answer 1

Answer:

We have S=S

3

+S

7

−S

21

,

S

3

= Sum of all the number between 200 and 500 which are divisible by 3

S

3

=201+204+...+498

201+[n−1]

3

=498⇒n=100

∴S

3

=

2

100

[201+498]=50×699=34950

S

7

=203+210+...497.

497=303+[n−1]7⇒n=43.

∴S

7

=

2

43

[201+497]=15050

S

21

=210+231+...+483

483=210+[n−1]21⇒n=14

∴S

21

=

2

14

[210+483]=4851

∴S=S

3

+S

7

−S

21

=34950+15050−4851=45149

Answer 2

Step-by-step explanation:

Correct option is

A

45149

We have S=S

3

+S

7

−S

21

,

S

3

= Sum of all the number between 200 and 500 which are divisible by 3

S

3

=201+204+...+498

201+[n−1]

3

=498⇒n=100

∴S

3

=

2

100

[201+498]=50×699=34950

S

7

=203+210+...497.

497=303+[n−1]7⇒n=43.

∴S

7

=

2

43

[201+497]=15050

S

21

=210+231+...+483

483=210+[n−1]21⇒n=14

∴S

21

=

2

14

[210+483]=4851

∴S=S

3

+S

7

−S

21

=34950+15050−4851=45149