Question

Find the sum of three digit number which are divisible by 11.

Answer 1

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 !!

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Answer 2

Answer:

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