A person is to count 4500 currency notes. Let a denote the number of notes he counts in the nth minute. If a1 = a2 = ... = a10 = 150 and a10, a11, ...are in an AP 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
tn = a + (n-1)d and Sn = n/2 [ 2a + (n-1)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 + (n-1) x -2] = 3000
⇒ n2-149n +3000 = 0
⇒ (n-24)n-125) = 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 + (125-1)(-2)
= 148 - 124 x 2 = 148-248 = -100
⇒ Number of notes cannot be negative.
