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Since he counts at constant rate of 150 notes per minute for first 10 minutes, He counts 150*10 = 1500 notes in that time.
So , the remaining notes are 4500 -1500 = 3000 notes.
a11= 150-2*1
a12= 150 -2*2
a13= 150 - 2*3
...
an= 150 -2*(n-10)
let n-10 =t;
a11+a12+a13+..an = 150*t - 2*(t(t+1))/2
Since LHS =3000
3000 = 150*t -(t*t+t)
3000 = 149*t - t*t
Which is a quadratic whose solution is t=24 or t=125 which translates to n=34 or n=135.
But we can't have n=135, because the terms goes negative and we cant have negative counting.
Hence, the answer is B. 34 minutes.

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