Question

Find the sum of all 3-digit natural numbers, which are multiples of 11.​

Answer 1

First three digit number divisible by 11 is 110

Last three digit numbers divisible by 11 is 990

∴a=110 a

n

=990 d=11 x=2

∴ No. of =2

Terms

a

n

=a+(n−1)d

990=110+(n−1).11

11

880

=x−1

80=x−1

x=81

∴ sum of all terms of A.P is given by

S

n

=[a+a

n

]×

2

n

=[110+990]×

2

81

=1100×

2

81

=550×81

=44550

let me follow u