A person is to count 4500 currency notes. Let $a_n$ denote the number of notes he counts in the $n^{\text{th}}$ minute. If $a_1=a_2=\ldots=a_{10}=150$ and $a_{10}, a_{11}, \ldots$ are in A.P. with common difference $-2$, then the time taken by him to count all notes is

JEE Main · 2010 · 11 Apr 2010 · Offline · Sequences and Series

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} = \dots = a_{10} = 150

and

a_{10}, a_{11}, \dots

are in A.P. with common difference

- 2

, then the time taken by him to count all notes is

Answer: (A)

In the first ten minutes he counts

10 \times 150 = 1500

notes, leaving

4500 - 1500 = 3000

. From the eleventh minute the counts form an arithmetic progression starting at

a_{11} = 148

with common difference

- 2

. Setting the sum of

n

such terms to

3000

\frac{n}{2} [2 (148) + (n - 1) (- 2)] = 3000

, which simplifies to

n^{2} - 149 n + 3000 = 0

, so

n = 24

n = 125

. The counts must stay positive, which rules out

125

, so

n = 24

and the total time is

10 + 24 = 34

minutes.

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option