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Start learning 50% faster. Sign in nowThe three-digit natural numbers start from 100 and end with 999. The first three-digit number which is divisible by 3 is 102. The last three-digit number which is divisible by 3 is 999. The numbers divisible by 3 form an arithmetic sequence with the first term a = 102 and the last term Tn = 999, and a common difference d = 3. Use the formula for the n-th term of an arithmetic sequence: Tn = a+(n-1) × d 999=102 + (n-1) × 3 999-102= (n-1) × 3 897 =(n-1) × 3 (n-1) =299 n =300
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