A person is to count $4500$ currency notes. Let an 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 the notes is

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$24$ minutes

$34$ minutes

$125$ minutes

$135$ minutes

answer is D.

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Detailed Solution

Suppose he takes $n$ minutes to count $4500$ notes. We have $a_{1} + a_{2} + . . . . + a_{10} = 10 (150), a_{11} = 148$ and $a_{11}, a_{12}, . . . . . . . a_{n}$ is an A.P. with common difference $d = - 2 .$
We are given
$\begin{array}{ll} a_{1} + a_{2} + \dots + a_{10} + a_{11} + \dots + a_{n} = 4500 \\ \Rightarrow a_{11} + a_{12} + \dots + a_{n} = 3000 \\ \Rightarrow \frac{n - 10}{2} [a_{11} + a_{n}] = 3000 \\ \Rightarrow \frac{1}{2} (n - 10) [148 + 148 + (n - 11) (- 2)] = 3000 \\ \Rightarrow (n - 10) (148 - n + 11) = 3000 \\ \Rightarrow (n - 10) (159 - n) = 3000 \\ \Rightarrow n^{2} - 169 n + 4590 = 0 \\ \Rightarrow n^{2} - 135 n - 34 n + 4590 = 0 \\ \Rightarrow (n - 135) (n - 34) = 0 \Rightarrow n = 135, 34 \end{array}$
All the notes get counted in $34$ minutes.