Question

find the sum of all 3digit number divisible hby 11​

Answer 1

Answer:

44550

Step-by-step explanation:

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Answer 2

Answer:

Step-by-step explanation:

the least 3 digit number is divisible by 11 is a= 110

the highest 3 digit number is divisble by 11 is l= 990

d=11

the last term is l=a+(n-1)d

990=110+(n-1)11

990=110+11n-11

990-110=11(n-1)

880=11(n-1)

n-1=880/11

n-1=80

n=80+1=81

n=81

the sum of the series is given as

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

81/2[2(110)+(81-1)11]

81/2[220+880]

81*550=44550