One of the most important and exam oriented question from Chapter name- Arithmetic Progression
Class 10th
Chapter number- 9
Exercise :- 9.6
This type of question has been asked in previous years exams.
In this question we have been asked to find the sum of all 3–digit natural numbers which are divisible by 13.
CBSE DHANPAT RAI publication
CBSE Mathematics Class 10th
Question 11(iii)
All 3–digit natural numbers which are divisible by 13 are 104, 117,…… ,988.
These numbers form an A.P. with first term(a) = 104 and
Common difference(d) = 117 − 104 = 13.
We know, the nth term of an A.P. id given by, an = a + (n − 1)d.
=> 988 = 104 + (n − 1)13
=> 988 = 104 + 13n -13
=> 988 = 91 + 13n
=> 13n = 897
=> n = 69
Also, we know sum of n terms of an A.P. is given by, Sn = n[2a + (n − 1)d] / 2.
S69 = 69[2(104) + (69 − 1)13]/2
= 69[1092]/2
= 69(546) = 37674
Hence, the sum of all 3–digit natural numbers which are divisible by 13 is 37674.