Question

Find the sum of all two digits numbers which are divisible by 11.​

Answer 1

Answer:

Answer. The sum of all 3 digits divisible by 11 is 44,550. To find 'the last term of a series' is = a + (n-1) d. Where a = 110, d = 11.

Answer 2

Given,

The numbers are two digits numbers which are divisible by 11.

To find,

The sum of those number.

Solution,

We can simply solve this mathematical problem by using the following mathematical process.

A two digit number which is divisible by 11, will have the same digit both in it's tens place and units place.

So, two digit numbers which are divisible by 11, are :

11,22,33,44,55,66,77,88,99

The above mentioned numbers form an AP series. So, we can apply the sum an AP formula instead of taking the sum manually.

First term of AP (a) = 11

Common difference (d) = 11

Number of terms (n) = 9

So,

Sum of AP = n/2 [2a+(n-1)×d]

= 9/2 [(2×11)+(9-1)×11]

= 9/2 (22+88)

= 9/2 × 110

= 495

Hence, the sum will be 495