Question

A person is to count 4500 currency notes. Let $a_n$​ 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 AP with a common difference $−2$, then the time taken by him to count all notes is?
A) 12.5 min
B) 135 min
C) 34 min
D) 24 min

Answer 1

Hint: Here, we are given 4500 currency notes and all the terms are in AP having a common difference 2. After that, we will find the notes counted till 10th min. then we will use the AP formula: \[{S_n} = \dfrac{n}{2}\{ 2a + (n - 1) \times d\} \] where, ‘a’ is the first term of the progression and ‘d’ is the difference between two consecutive terms and Sn is the sum of the first n terms. Thus, we will use the factorization method to solve this and get the value of n.

Complete step by step solution:
Given that, There are a total of 4500 notes.
\[ \Rightarrow {a_1} + {a_2} + {a_3} + .... + {a_n} = 4500\]
Also, given that, \[{a_1} = {a_2} = ..... = {a_{10}} = 150\]
So, the number of notes counted in first 10 minutes
\[ = 150 \times 10\]
\[ = 1500\] notes
Thus, till the 10th minute, the number of notes counted = 1500.
Suppose, that the person counts the remaining 3000 (= 4500 - 1500) currency notes in $n$ minutes.
Given that, $a_10, a_11, …..$ are in AP and also there is a common difference between them is $2$.
3000 = Sum of n terms of an A.P. with first term 148
\[ \Rightarrow 3000 = \dfrac{n}{2}\{ 2 \times 148 + (n - 1) \times ( - 2)\} \]
Simplify this, we will get,
\[ \Rightarrow 3000 = \dfrac{n}{2}\{ 2(148) - 2(n - 1)\} \]
\[ \Rightarrow 3000 = n\{ 148 - (n - 1)\} \]
Removing the brackets, we will get,
\[ \Rightarrow 3000 = n\{ 148 - n + 1\} \]
\[ \Rightarrow 3000 = n\{ 149 - n\} \]
\[ \Rightarrow 3000 = 149n - {n^2}\]
By using transposing and moving the RHS terms to LHS, we will get,
\[ \Rightarrow 3000 - 149n + {n^2} = 0\]
\[ \Rightarrow {n^2} - 149n + 3000 = 0\]
By using factorization method, we will get,
\[ \Rightarrow {n^2} - 125n - 24n + 3000 = 0\]
\[ \Rightarrow n(n - 125) - 24(n - 125) = 0\]
\[ \Rightarrow (n - 125)(n - 24) = 0\]
\[ \Rightarrow n - 125 = 0\] or \[n - 24 = 0\]
\[ \Rightarrow n = 125\] or \[n = 24\]
For n = 125
\[ \Rightarrow n = 125 + 10 = 135\]
\[\therefore {a_{135}}\]
\[ = 148 + (135 - 1)( - 2)\]
\[ = 148 - 268 < 0\]
Here, n = 125 is not possible.
Next, n = 24
\[ \Rightarrow n = 24 + 10 = 34\]
Therefore, the time taken by him to count all notes is 34 mins. So, option (C) is correct.

Note:
Alternative method:
\[{a_1} + {a_2} + {a_3} + .... + {a_n} = 4500\]
\[ \Rightarrow {a_{11}} + {a_{12}} + .... + {a_n} = 4500 - (10 \times 150)\]
\[ \Rightarrow {a_{11}} + {a_{12}} + .... + {a_n} = 3000\]
\[ \Rightarrow 148 + 146 + .... = 3000\]
\[ \Rightarrow \dfrac{{n - 10}}{2}(2 \times 148 + (n - 10 - 1)( - 2)) = 3000\]
Let m = n – 10 and applying this, we will get,
\[ \Rightarrow \dfrac{m}{2}(2 \times 148 + (m - 1)( - 2)) = 3000\]
\[ \Rightarrow m(148 - (m - 1)) = 3000\]
\[ \Rightarrow 148m - m(m - 1) = 3000\]
\[ \Rightarrow 148m - {m^2} + m = 3000\]
\[ \Rightarrow 149m - {m^2} = 3000\]
\[ \Rightarrow 0 = 3000 - 149m + {m^2}\]
\[ \Rightarrow {m^2} - 149m + 3000 = 0\]

Solving this, we will get m = 24 or m = 125.
Then, we will get, n = 34 or n = 135

Hence, the time taken by the person to count all the notes is 34 min.

An arithmetic progression (AP) is a sequence of numbers such that the difference of any two successive members is a constant. The common difference is denoted by d. Also, the sum of the first n elements of the arithmetic series is denoted by Sn. Arithmetic series means the sum of the elements of an arithmetic progression. Even in the case of odd numbers and even numbers, we can see the common difference between two successive terms will be equal to 2.