A person is to count 4500 currency notes. Let a_n denote the number of notes he counts in the n^th 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

First 10 minutes = 150 x 10 = 1500 148 + 146 + 144 + * s 3000 = (296n - 2n^2 + 2n / 2) 4500 = 1500 + (n / 2) ( 296 + (n - 1)(-2) ) 2n^2 + 298n + 6000 = 0 n^2 - 149n + 3000 = 0 n^2 - (125 + 24)n + 125 * 24 = 0 (n - 24)(n - 125) = 0 n = 24 or n = 125 n eq 125 => n = 24 Total time = 24 + 10 = 34 minutes

A person is to count 4500 currency notes. Let a_{n} de... | Sequence & Series (Mathematics)

\text{First 10 minutes} = 150 \times 10 = 1500

148 + 146 + 144 + \cdots

3000 = \frac{296n - 2n^2 + 2n}{2}

4500 = 1500 + \frac{n}{2} \left( 296 + (n - 1)(-2) \right)

2n^2 + 298n + 6000 = 0

n^2 - 149n + 3000 = 0

n^2 - (125 + 24)n + 125 \cdot 24 = 0

(n - 24)(n - 125) = 0

n = 24 \quad \text{or} \quad n = 125

n eq 125 \Rightarrow n = 24

\text{Total time} = 24 + 10 = 34 \text{ minutes}

Explanation

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