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Permutations And Combinations
A person is to count 4500 currency notes. Let a denote the number of notes he counts in the nth minute. If a_{1} = a_{2} = ... = a_{10} = 150 and a_{10}, a_{11}, ...are in an A_{P} with common difference 2, then the time taken by him to count all notes is
24 min
34 min
125 min
135 min
B.
34 min
Let the first term of an AP be a and common difference be d and number of terms be n, then
t_{n} = a + (n1)d and Sn = n/2 [ 2a + (n1)d]
Number of notes that the person counts in 10 min = 10 x 150 = 1500
Since a10, a11, a12, .... are in AP in with common difference 2
Let n be the time has taken to count remaining 3000 notes than
n/2[2 x 148 + (n1) x 2] = 3000
⇒ n2149n +3000 = 0
⇒ (n24)n125) = 0
n = 24, 125
Then, the total time taken by the person to count all notes = 10 +24 = 34 min
n = 125 is discarded as putting n = 125
an = 148 + (1251)(2)
= 148  124 x 2 = 148248 = 100
⇒ Number of notes cannot be negative.
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