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Let the time be = t hrs , speed be= 48 km/hr , distance = 48× t Distance travelled by B forms an AP = 40, 40½, 41, 41½ , ………………………. t terms Where First term is 40 and Common difference is 1/2 Now, Sum of n terms = Sn = n/2 [2a + (n-1)d] = distance covered by A Sn = t/2 [2 × 40 + (t-1) 1/2] 48t= t/2 [2 × 40 + (t-1) 1/2] 96 = 80 + (t – 1) 1/2 16 = (t – 1) 1/2 33 = t Therefore A and B will meet after 33 hours
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