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1. Chapter 9 Class 11 Sequences and Series
2. Serial order wise
3. Ex 9.2

Transcript

Ex 9.2,2 Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5. Multiples of 5 are 5, 10, 15,20,25 …… Multiples of 5 between 100 and 1000 are 105, 110,115, … ,990,995. This sequence forms an A.P. as difference between the consecutive terms is constant. Here, first term = a = 105 Common difference = d = 110 – 105 = 5 & last term = l = 995 First we need to find number of terms, i.e. n 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 = 995 , a = 105 , d = 5 Putting values 995 = 105 + (n – 1)5 995 = 105 + 5n – 5 995 = 105 + 5n – 5 995 = 105 – 5 + 5n 995 = 100 + 5n 995 - 100 = 5n 895 = 5n 895/5 = n 179 = n n = 179 For finding sum, we use the formula Sn = n/2 [2a + (n – 1)d] Putting n = 179 , a = 105 , d = 5 = 179/2 [2 × 105 + (179 – 1)(5)] = 179/2 [2 × 105 + (179 – 1)(5)] =179/2[210 + (178)5] = 179/2 [210 + 890] = 179/2  = 179 × 550 = 98450 Hence the sum of all natural number lying between 100 & 1000 which are multiple of 5 is 98450

Ex 9.2 