This is the basic and conceptual question from Chapter name- Arithmetic Progression

Chapter number- 9

Exercise 9.6

In this question we have to find the sum Of all 3–digit natural numbers which are multiples of 11.

CBSE DHANPAT RAI publications

Class:- 10th

Solutions of CBSE Mathematics

Question 12(iv)

All 3–digit natural numbers which are divisible by 11 are 110, 121, 132,…… ,990.

These numbers form an A.P. with first term(a) = 110 and

Common difference(d) = 121 − 110 = 11.

We know, the nth term of an A.P. id given by, a

_{n}= a + (n − 1)d.=> 990 = 110 + (n − 1)11

=> 990 = 110 + 11n -11

=> 990 = 99 + 11n

=> 11n = 891

=> n = 81

Also, we know sum of n terms of an A.P. is given by, S

_{n}= n[2a + (n − 1)d] / 2.S

_{81}= 81[2(110) + (81 − 1)11]/2= 81[1100]/2

= 81(550) = 44550

Hence, the sum of all 3–digit natural numbers which are divisible by 13 is 44550.