Question
The sum of all natural numbers between 100 and 200, which are multiples of 3 is :
Answer: Option C
Answer: (c)Numbers divisible by 3 and lying between 100 and 200 are : 102, 105,..... 198 Let number of terms = n198 = 102 + (n–1) 3n–1 = ${198 - 102}/3$ = 32n = 33S = $n/2$(a+l) =$33/2$(102+ 198) = 4950
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Answer: (c)Numbers divisible by 3 and lying between 100 and 200 are : 102, 105,..... 198 Let number of terms = n198 = 102 + (n–1) 3n–1 = ${198 - 102}/3$ = 32n = 33S = $n/2$(a+l) =$33/2$(102+ 198) = 4950
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