Question

what is the sum of all natural numbers 100 to 250 divisible by 6​

Answer 1

Step-by-step explanation:

Integers between 100 and 200 divisible by6 are-

102,108,114,...198

Here a=102,d=6 last term l=198

t

n

=a+(n−1)d

⇒198=102+(n−1)6

⇒(n−1)6=198−102

=96

⇒(n−1)6=96⇒n−1=16

n=17

∴Sum=

2

n

[a+l]=

2

17

[102+198]

=

2

17

×300=150×17=2550