Arithmetic Progression
Question 2 Deleted for CBSE Board 2025 Exams You are here
Question 3 Important Deleted for CBSE Board 2025 Exams
Question 4 Deleted for CBSE Board 2025 Exams
Question 5 Important Deleted for CBSE Board 2025 Exams
Question 6 Deleted for CBSE Board 2025 Exams
Question 7 Important Deleted for CBSE Board 2025 Exams
Question 8 Deleted for CBSE Board 2025 Exams
Question 9 Important Deleted for CBSE Board 2025 Exams
Question 10 Deleted for CBSE Board 2025 Exams
Question 11 Important Deleted for CBSE Board 2025 Exams
Question 12 Deleted for CBSE Board 2025 Exams
Question 13 Deleted for CBSE Board 2025 Exams
Question 14 Important Deleted for CBSE Board 2025 Exams
Question 15 Important Deleted for CBSE Board 2025 Exams
Question 16 Important Deleted for CBSE Board 2025 Exams
Question 17 Deleted for CBSE Board 2025 Exams
Question 18 Important Deleted for CBSE Board 2025 Exams
Arithmetic Progression
Last updated at April 16, 2024 by Teachoo
Question2 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 [1100] = 179 × 550 = 98450 Hence the sum of all natural number lying between 100 & 1000 which are multiple of 5 is 98450