Question

    Find the sum of all those 3-digit numbers which when

    divided by 10 leave a remainder of 5.
    A 49500 Correct Answer Incorrect Answer
    B 72000 Correct Answer Incorrect Answer
    C 56500 Correct Answer Incorrect Answer
    D 54800 Correct Answer Incorrect Answer

    Solution

    Numbers 105, 115, 125, ……985, 995 first number (a) = 105 The last number (8) = 995 All difference (d) =10 From the formula, nth term (Tn) = a + (n-1) d, 995=105+(n-1) 10   n = 90   Sum of n terms (Sn) =n/2 [2a+(n-1) d]   90/2[2 ×105+ (90-1)10] = 45 [210+890] = 45 × 1100   = 49500.

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