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Last updated at Dec. 8, 2016 by Teachoo

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Misc 5 Find the sum of integers from 1 to 100 that are divisible by 2 or 5. Sum of integers divisible by 2 or 5 = Sum of integers divisible by 2 + Sum of integers divisible by 5 – Sum of integers divisible by 2 & 5 Finding sum of numbers from 1 to 100 divisible by 2 Integers divisible by 2 between 1 to 100 are 2, 4, 6, 8, …100 This forms an A.P. as difference of consecutive terms is constant. First term = a = 2 common difference d = 4 – 2 Last term = l = 100 2, 4, 6, 8, …100 Number of terms = n = (100 )/2 = 50 For finding sum, we use the formula Sn = n/2 [a + l] Here, n = 50 , l = 100 & a = 2 Sn = 50/2 (2 + 100) = 25 × 102 = 2550 Hence, sum of numbers from 1 to 100 divisible by 2 is 2550 Finding sum of numbers from 1 to 100 divisible by 5 Integer between 1 to 100 which are divisible by 5 are 5, 10, 15, 20, 25, 30, … 100 This forms an A.P. as difference of consecutive terms is constant Here, First term a = 5 Common difference d = 10 – 5 Last term l = 100 First we calculate number of terms in this AP We know that an = a + (n – 1)d where an = nth term , n = number of terms, a = first term , d = common difference Here, an = last term = l = 100, a = 5 , d = 5 Putting values 100 = 5 + (n – 1) 5 100 – 5 = (n – 1)5 95 = (n – 1)5 95/5 = (n – 1) 19 = n – 1 19 + 1 = n 20 = n n = 20 For finding sum, we use the formula Sn = n/2 [a + l] Here, n = 20 , l = 100 & a = 5 = 20/2 ( 5 + 100) =10 (105) = 1050 Hence, sum of numbers from 1 to 100 divisible by 5 is 1050 Finding sum of numbers from 1 to 100 divisible by 2 & 5 both Integers between 1 to 100 which are divisible by both 2 and 5 are 10, 20, 30,… ,90,100. This forms an A.P. as difference of consecutive terms is constant Here, First term = a = 10 Common difference d = 20 – 10 Last term = l = 100 & Number of terms = n = 10 For finding sum, we use the formula Sn = n/2 [a + l] Here, n = 20 , l = 100 & a = 5 S10 = 10/2 ( 10 + 100) = 5 (110) = 550 Hence, sum of numbers from 1 to 100 divisible by 5 & 2 both is 550 Now finding sum of integers from 1 to 100 which are divisibly by 2 or 5 Sum of integers divisible by 2 or 5 = Sum of integers divisible by 2 + Sum of integers divisible by 5 – Sum of integers divisible by 2 & 5 = 2550 + 1050 – 550 = 3050 Thus, sum of integers from 1 to 100 which are divisible by 2 or 5 is 3050

Miscellaneous

Misc 1

Misc 2

Misc 3 Important

Misc 4

Misc 5 You are here

Misc 6

Misc 7 Important

Misc 8

Misc 9

Misc 10

Misc 11

Misc 12

Misc 13

Misc 14

Misc 15

Misc 16 Important

Misc 17

Misc 18

Misc 19 Important

Misc 20

Misc 21

Misc 22

Misc 23

Misc 24

Misc 25 Important

Misc 26

Misc 27

Misc 28 Important

Misc 29

Misc 30

Misc 31

Misc 32 Important

Chapter 9 Class 11 Sequences and Series

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.