The sumof all three digit integers are divisible by 9

17 and 19

Step-by-step explanation:

9+1+7= 17 and 9+2+8=19

Heya user ,

Here is your answer !!

The first 3-digit number divisible by 11 is 110 .

The last 3-digit number divisible by 11 is 990 .

So , the A.P. series is 110 , 121 , 990 .

Common difference = 11 .

an = a1 + ( n - 1 ) d

=> 990 = 110 + ( n - 1 ) 11

=> 880 = ( n - 1 ) 11

=> 80 = ( n - 1 )

=> 81 = n .

So , number of 3 digit numbers divisible by 11 is 81 .

Now , the middlemost term is the 41th term .

So , a41 = a1 + ( n - 1 ) d [ where n = 41 ]

=> a41 = 110 + (41-1)*11

=> a41 = 110 + ( 11*40 )

=> a41 = 110 + 440

=> a41 = 550 .

So , the middlemost term of this A.P. series is 550 .

Hope it helps !!

Please mark it as brainlist

STEP BY STEP EXPLANATION

a=110

d=11

an=990

n=?

A.T.Q

an=a+(n-1)d

990 =110+(n-1)11

990-110=(n-1)11

880=(n-1)11

88+1=n

89=n

Sn=n/2[2a+(n-1)d]

=89/2[2*110+(88)11]

=89/2[220+968]

=89/2[1188]

=89*594

=52866

answer:

no book sorry.. pls post a photo along with question