Question

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 )

# 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

Solution