Question

find the sum of all three digit natural number which are multiple of 11

Answer 1

---------------------------------------------------------------------------------------------------------

We know that the first 3 digit natural number that is divisible by 11 is 110 and the last 3 digit natural number that is multiple of 11 is 990.All the multiples of 11 will form an A.P that's  common difference will be 11.

So,

First term = 110.

Last term = 990.

Number of terms = ?

Sum of all terms = ?

Let numbers of terms are ' n '.

Nth term = First term + ( number of terms - 1 ) common difference

990 = 110 + ( n - 1 ) 11

990 = 110 + 11n - 11

11n = 990 - 110 + 11

11n = 891

n = 891 / 11

n = 81.


Sum of all terms of an A.P = n/2 { First term + Last term )


                                          = 81 / 2 { 110 + 990 }

                                          = 81 / 2 *  1,100

                                          = 81 *  550

                                          = 44,550

Sum of all 3 digits natural number that is divisible by 11 is 44,550.

------------------------------------------------------------------------------------------------------