Question

The sum of integers lie between 1 and 121
(both included) and divisible by 3 or 5 or
11 is...
​

Answer 1

Answer:

Step-by-step explanation: the integers that lie betweem 1 and 121 and are divisible by 3,5,11 are

#by 3 are

3 x1

3x2 .........

3x40

sum of the numbers in AP with d =3 ,a=3,n=40,l=120

sum= 40/2 [3+120] =20 x 123 =2460

#by 5 are

5x1

5x2.......

5x 24

sum of the numbers in ap with d=5,a=5,l=120,n=24

sum = 24/2 [5+120] =1500

#by 15 are

numbers which are divsible by both 5 and 3 = 15,30,30,45,60,75,90,105,120

sum of numbers in ap with a=15,d=15,n=8,l=120

sum = 8/2 [15+120]=540

# by 11 are

11x1

11x2......

11x10

sum of the numbers in ap with n=10.a=11,d=11,l=110

sum = 10/2[11+110]=605

the final answer is = 2460+1500+605-540= 4025

pls mark this answer as the brainiest answer.