A person is to count 4500 notes. Let a_n denote the number of notes he counts in the nth minute. If a₁=a₂=a₃= . =a₁₀=150 , and a₁₀, a₁₁, a₁₂, . are in AP with the common difference -2 , then the time taken by him to count all the notes is

Step-by-step solution

Given, a ₁= a ₂= a ₃= =a₁₀=150 Also, a ₁₀, a ₁₁, a ₁₂, . are in A.P. and d =-2 Since, a ₁₀=150, ~A . P . is 150,148,146, . For the first 10 minutes, he has counted 150 10=1500 notes. Time taken to count remaining 3000 notes aligned & S _ n = n 2 [2 a +( n -1) d ] & 3000= n 2 [2 148+( n -1)(-2)] aligned aligned & 3000= n 2 2(148- n +1) & 3000=148 n - n ^2+ n & n ^2-149 n +3000=0 aligned aligned & ( n -24)( n -125) & n =24, or n =125 . aligned Since he has taken 10 minutes to count 1500 notes, he will not take 125 mi