Q. 173.7( 9 Votes )

Find the sum of all three - digit natural numbers which are divisible by 13.

Answer :

Three - digit natural numbers which are divisible by 13 are 104, 117, 130, …, 988.

Sum of these numbers forms an arithmetic series 104 + 117 + 130 + … + 988.


Here, first term = a = 104


Common difference = d = 13


We first find the number of terms in the series.


Here, last term = l = 988


988 = a + (n - 1)d


988 = 104 + (n - 1)13


988 - 104 = 13n - 13


884 = 13n - 13


884 + 13 = 13n


13n = 897


n = 69


Now, Sum of n terms of this arithmetic series is given by:


Sn = [2a + (n - 1)d]


Therefore sum of 69 terms of this arithmetic series is given by:


S69 = [2(104) + (69 - 1)(13)]


= (69/2) × [208 + 884]


= (69/2) × 1092


= 69 × 546


= 3767


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Champ Quiz | Arithmetic Progression34 mins
Champ Quiz | Arithmetic Progression30 mins
Lets Check Your Knowledge in A.P.49 mins
Arithmetic progression: Previous Year NTSE Questions34 mins
Quiz on Arithmetic Progression Quiz32 mins
Become a Master of A.P. in 45 Minutes!!47 mins
Arithmetic Progression Tricks and QUIZ37 mins
Quiz | Group of Questions on General Term of an A.P49 mins
Get to Know About Geometric Progression41 mins
NCERT | Solving Questions on Introduction of A.P42 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses