Find the sum of all those 3-digit numbers which when divided by 10 leave a remainder of 5.
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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