Question

Find the sum of all natural numbers between 250 and 1000 which are divisible by 9. (Using Arithmetic Progression). [Ans. = 52542]

Answer 1

hope it helps..............

Answer 2

Hey !!
=========

Here is your answer !!!
===================

Natural Number from 250 to 1000 divisible by 9
Then
First term = 252 ( divisible by 9 on 28 times and is the first term from 250 which is divisible by 9 )

last term = 999

Now!!

999 = 252 + ( n - 1 ) d

Where d is the common difference which will obviously be equal to
999 = 252 + ( n - 1 ) 9
999 = 252 + 9n - 9
999 = 243 + 9n
756 = 9n
n = 84

Sum !!

Sn = n / 2 ( 2a + ( n - 1 ) d
Sn = 84/2 ( 2 × 252 + ( 84 - 1 ) 9 )
Sn = 42 ( 504 + 83 × 9)
Sn = 42 ( 504 + 747)
Sn = 42 × 1251
Sn = 52542

Hence the sum = 52542

Hope it helped you !!

Plz Mark Brainliest !!