Give a 3-digit number divisible 9 and 11 and the sum of the digits is 18.
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Answer:
The least 3 digit number that is divisible by 11 is a=110
The highest 3 digit number that is divisible by 11 is l=990
d=11
Last term is l=a+(n−1)d
990=110+(n−1)11
880=(n−1)11
n−1=
11
880
n−1=80⇒n=81
The Sum of the series is given as
S
n
=
2
n
[2a+(n−1)d]
S
81
=
2
81
[2(110)+(81−1)11]
=
2
81
[220+880]
=81×550=44550
Step-by-step explanation:
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