Question

Find the sum of all integers between 100 and 1000 which are divisible by9​

Answer 1

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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 !!