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Misc 6 - Find sum of all two digit numbers which divided by 4 - Miscellaneous

Misc 6 - Chapter 9 Class 11 Sequences and Series - Part 2
Misc 6 - Chapter 9 Class 11 Sequences and Series - Part 3
Misc 6 - Chapter 9 Class 11 Sequences and Series - Part 4
Misc 6 - Chapter 9 Class 11 Sequences and Series - Part 5

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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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.