Find the sum of all natural numbers lying between 100 and 1000 , which are multiples of 5.

Answer

Related Courses

Precalculus

NCERT Class 11 - Math

Chapter 9

Sequences and Series

Section 2

Sequences

Discussion

You must be signed in to discuss.

Top Precalculus Educators

Watch More Solved Questions in Chapter 9

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Video Transcript

well we have problem number two in which we need to find all the natural numbers lying between 101,000. So this is 100 And this is 1000 which are multiples of five. So we have to exclude 100. So this will be 105. and of course 995. So we have to find some of this some of these numbers. So this is the first term. And Accommodation will be obviously five. Sorry, common difference will be obviously five. So this is an a thematic sequence. And last term is 995. So let us find the value of n. Okay, so we have L equal to a plus And -1 in today. From here we can find the value of NSL minus a by d plus one. So Early in the last term, which is 995. So let us find the value of n 995. -1 term is 105, divided by five Plus 1. So this is 890 By five plus 1. Yeah. So Okay, okay, 178 Plus one. This is the value of and so as and we have As an equal to and by two April cell. So as 179 equal to 179 x two, It is 105 plus 1995. So this is 179 x two and uh zero zero. So it is 1100 550, 179 into 550, 179 and 2550 98,000 98,450 should be the required answer. Thank you so much.

Get More Help with this Textbook