Question

Find the sum of all 3-digit natural numbers, which are multiples of 11. (CBSE 2012)​

Answer 1

Answer:

44550

Step-by-step explanation:

So, the sequence of three digit numbers which are divisible by 11 are 110, 121, 132, …, 990. Clearly, it is an A.P. with first term, a = 110 and common difference, d = 11. Let there be n terms in the sequence. Hence, the sum of all three digit numbers which are multiples of 11 is 44550

Answer 2

Answer:

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