Question

Find the sum of all natural numbers between 300 and 500 which are divisible by 11.

Answer 1

Hey,sup!

We'll solve this using Arithmetic Progression.

As per the question,

a=308. (First number divisible by 11 in the range)
d= 11.
l=495. (Last number divisible by 11 in the range)

A(n)= a+(n-1)d.

=>495=308+(n-1)11.

=>495-308=(n-1)11.

=>187=(n-1)11.

=>n-1= 187/11= 17.

=> n= 17+1 =18.

S(n)= n/2 (a+l).
S(18)= 18/2(308+495).
= 9 × 803.
= 7227.

Sum of all natural numbers between 300 and 500 which are divisible by 11 is 7227.

Hope it helps.

Answer 2

a=308

d=11

n=?

an=495

To find the number of terms ,

• an=a+(n-1)d
• 495=308+(n-1)11
• 495-308=(n-1)11
• 187=(n-1)11
• 187/11=(n-1)
• 17=n-1
• -n=-17-1
• -n=-18
• n=18

To find sum of the numbers,

• Sn=n/2(a+l)
• Sn=18/2(308+495)
• Sn=9×803
• Sn=7,227

Therefore,the sum of natural numbers between 300 and 500 which is divisible by 11 is 7,227..

HOPE IT IS HELPFUL..