Question

7. Find the sum of three digits numbers which are divisible by 11​

Answer 1

hi mate

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

2/n [2a+(n−1)d]

2/81 [2(110)+(81−1)11]

2/81 [220+880]

81×550=44550

i hope it helpfull to you

Answer 2

A = 110 ,D = 11 L=990 N= ?

Tn=A+(n-1) d

990= 110 + (n-1) 11

880 =(n-1) 11

880/11=(n-1)

80=n-1

81=n

Sn=n/2(2a+(n-1)d)

= 81/2 ( 2 [110] + 80 × 11)

= 81/2 (220+880)

=81/2 ( 1100 )

=81 × 550

=44,550

Therefore, the sum of numbers divisible by 11 that are three digit is 44,550

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