Last updated at May 29, 2018 by Teachoo

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Misc 6 Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder. Two digit numbers are 10,11,12,13, .98,99 Finding minimum number in 10,11,12,13, .98,99 which when divided by 4 yields 1 as remainder 10/4 = 22/4 11/4 = 23/4 12/4 = 3 13/4 = 31/4 So the sequence will start from 13 Finding maximum number in 10,11,12,13, .98,99 which when divided by 4 yields 1 as remainder 99/4 = 243/4 98/4 = 242/4 97/4 = 241/4 So the sequence will end at 97 Thus, the sequence starts with 13 and ends with 97 Thus, the two digit numbers which are divisible by 4 yield 1 as remainder are 13, 17, 21, 25, 93,97. This forms an A.P. as difference of consecutive terms is constant. 13, 17, 21, 25, 93,97. First term a = 13 Common difference d = 17 13 = 4 Last term l = 97 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 = 97 , a = 13 , d = 4 Putting values 97 = 13 + (n 1)4 97 13 = (n 1)4 84 = (n 1)4 84/4 = (n 1) 21 = n 1 21 + 1 = n 22 = n n = 22 For finding sum, we use the formula Sn = n/2 [a + l] Here, n = 22 , l = 97 & a = 13 S22 = 22/2 [13 + 97] = 11 [110] = 1210 Hence, sum of all two digit numbers which when divided by 4, yields 1 as remainder is 1210

Miscellaneous

Misc 1

Misc 2

Misc 3 Important

Misc 4

Misc 5

Misc 6 You are here

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

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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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.