Question

find the sum of all integers between 100 and 1000 which are divisible by 9

Answer 1

the sum of no. will be divisible by 9 for example
we will take 729 as a no. to check the divisivlity of 9
the sum of no. is 7+2+9= 18
it is divisible by 9 so the no. will also be divisible by 9

Answer 2

Hey there !

Solution:

This is a sequence and series based sum. Hence we are using the concept of Arithmetic Progression for solving it.

First Term ( a ) = 108

Common Difference ( d ) = 9

Number of terms = ?

Last term ( l ) = 999

So Using the formula below, we get,

l = a + ( n - 1 ) d

=> 999 = 108 + ( n - 1 ) 9

=> 999 - 108 = 9 ( n - 1 )

=> 891 = 9 ( n - 1 )

=> 891 / 9 = ( n - 1 )

=> 99 = ( n - 1 )

=> n = 99 + 1 = 100 terms

Hence Sum of all terms can be found out by applying the formula:

S ( n ) = n / 2 [ a + l ]

=> S ( 100 ) = 100 / 2 [ 108 + 999 ]

=> S ( 100 ) = 50 ( 1107 )

=> S ( 100 ) = 50058

Hence the sum of all integers divisible by 9 present between 100 and 1000 is 50058.

Hope my answer helped !