Question

Sum of all numbers from 100 to 999, divisible by 11

Answer 1

@dpasingh

880/11=80

Answer 2

Oh yes sorry

Answer 3

\[81\times\frac{110+990}{2}=44550\]

Answer 4

The numbers from 100 to 999 divisible by 11 are as given below..
110, 121, 132,.......990
Therefore here first term a= 110, commomn difference d=11 and tn =990
Now we shall find the number of terms n
since tn = a+(n-1)d
i.e. 990= 110+(n-1)11
i.e. 990- 110=(n-1)11
i.e. 880=(n-1)11
i.e. 80=n-1
n= 81
Now sum of given numbers is given by
Sn = n/2[a+tn]
i.e. S81= 81/2[110+990]= 81/2[1100]=81*550=44550
i.e. S81=44550

Answer 5

Tyvm :)