A person is to count 4500 currency notes. Let a, denotes the number of notes he counts in the nth min. Ifa, = a, = 1 a2 ....... = 210 = 150 and 210, 211, ....... are in A.P with common difference -2, then the time taken = by him to count all notes is (A) 24 min (B) 34 min (C) 125 min (D) 135 min

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Norw no. of and es left to be counted = 3000 8 Uslng formula [ S n=frac{n}{2}[2 a+(n-1) d] ] Here ( mid begin{array}{c}a=150000 d=-2 quad s h=3000end{array} ) ( left.3000=frac{n}{2}[2[1500]+(n-1) in 2)right] ) [ 6000=n quad[3000-2 n+2] ] ( 6000=n[3082-2 n] ) ( 6000=308 pi 2 n-2 n^{2} ) 00 ( 2 n^{2}-30082 n+6000=0 ) Using quadratic formula ( frac{1500+frac{1}{6}(2,2000)}{2} frac{151 pm sqrt{10801}}{2} ) [ h=23 cdot 53 mathrm{mPns} ] but since we need to include initial loming so final time e 2 sands is and r ( 3 t ) ( left.bar{q}^{prime} mright|^{prime} r y )

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