Question

the sum of all the natural numbers between 1 and 101 which are divisible by 5 is

Answer 1

Answer:

1050.

Step-by-step explanation:

Numbers between 1 and 101 which are divisible by 5 are 5 , 10 , 15 ... 100.

On observing we found that these consecutive multiples of 5 form an AP of common difference 5 with first term of 5 and last term of 100.

From the properties of AP :

   nth term = a + ( n - 1 )d         { where a is the first term and d is the common difference between the terms }

Thus, let the number of multiples of 5 be n.

⇒ 100 = 5 + ( n - 1 )5

⇒ 100 - 5 = ( n - 1 )5

⇒ 95 = ( n - 1 )5

⇒ 95 / 5 = n - 1

⇒ 19 = n - 1

⇒ 20 = n

Therefore, using Sum of n terms = ( n / 2 ){ a + l }     where l is the last term

⇒ Sum of first 20 term = ( 20 / 2 )[ 5 + 100 ]

                ⇒ 10( 105 )

                ⇒ 1050

Hence the sum of all natural numbers between 1 and 101 which are divisible by 5 is 1050.

Answer 2

Step-by-step explanation:

Aloha !

$\text{ This is Twilight Astro ^_^}$

$Your answers:-$

To find n th term we have formula in progressions.

N th term= a+(n-1)d

100=5+(n-1)5

100-5 =(n-1)5

95=(n-1)5

95÷5=(n-1)

19=n-1

n=19+1

n=20

We need sum of n th term= n/2(a+I)

=20/2×(5+100)

=10/1×(105)

=10×105

Thank you

@ Twilight Astro ✌️☺️♥️