Question

find the sum of all natural numbers between 300 and 1000 which are divisible by 9
​

Answer 1

Step-by-step explanation:

Hope u got it frnds or guys..

Answer 2

Step-by-step explanation:

First: the first multiple of 7 greater or equal to 300 is 4*70 + 21 = 280 + 21 = 301

That is 7*(40+3) = 43*7

Last: the last one would be 490+7 = 497.

That is 7*(70+1) = 71*7

The number of terms is the amount of numbers from 43 to 71, including both.

That is: (71–43)+1 = 29 terms.

(29*7 = 203)

Then:

SumAP = 29*(301 + 497)/2 = 29*798/2 = 29*399 = 29*400 - 29 = 11600–29 = 11571