Question

A person is to count $4500$ currency notes. Let ${a_n}$ denotes the number of notes he counts in the nth minute and if ${a_1} = {a_2} = ......... = {a_{10}} = 150$ and ${a_{10}},{a_{11}}.......$ are all in AP with the common difference $ - 2$ then the time taken by him to count all the notes is:

Answer 1

Hint:
Here ${a_n}$ denotes the number of notes so ${a_1}$ contains $150$ notes and similarly ${a_2}.....{a_{10}}$ all also contains $150$ notes and this means that in $10$ minutes he counts $1500$ notes. Let $a = {a_{11}} = 148$ and the common difference is $ - 2$ and let the number of terms be n and we need to count the rest of the $3000$ notes. We can us the formula for sum of the AP
Sum of n terms $ = \dfrac{n}{2}(2a + (n - 1)d)$

Complete step by step solution:
Here in the question it is given that a person is to count $4500$ currency notes and ${a_n}$ denotes the number of notes he counts in the nth minute and also given that ${a_1} = {a_2} = ......... = {a_{10}} = 150$ and ${a_{10}},{a_{11}}.......$ are all in AP with the common difference $ - 2$ then we need to find time taken by him to count all the notes.
So here \[{a_1}\] means that the number of notes he can count in the first minute which is $150$
Similarly in the second minute also he is able to count the $150$ notes
So as it is given that ${a_1} = {a_2} = ......... = {a_{10}} = 150$
So in the first $10$ minutes he is able to count the $150(10) = 1500$ notes
Now we are given that he is to count the $4500$ currency notes so the number of notes left to be counted is $4500 - 1500 = 3000$
This means that $3000$ notes are still remaining and also we are given that ${a_{10}},{a_{11}}.......$ are all in AP with the common difference $ - 2$
So ${a_{11}} = {a_{10}} + d = 150 - 2 = 148$
Now let us assume that number of minutes be n and that means that the number of terms in AP be n
So for the remaining $3000$ notes we can use the sum of the AP formula
Sum of the n terms $ = \dfrac{n}{2}(2a + (n - 1)d)$
$\Rightarrow 3000 = \dfrac{n}{2}(2(148) + (n - 1)( - 2))$
$\Rightarrow 6000 = (196 - 2n + 2)n$
$\Rightarrow 6000 = 298n - 2{n^2}$
So we can write it as
$\Rightarrow 2{n^2} - 298n + 6000 = 0$
Taking $2$ common we get that
$\Rightarrow {n^2} - 149n + 3000 = 0$
Now we can split it as
$\Rightarrow {n^2} - 24n - 125n + 3000 = 0$
$\Rightarrow n(n - 24) - 125(n - 24) = 0$
$\Rightarrow (n - 24)(n - 125) = 0$
$\Rightarrow n = 125,24$
Now we get the two values of n
Now if we take$n = 125$
$\Rightarrow {a_n} = a + (n - 1)d$
$\Rightarrow {a_n} = 148 + (125 - 1)( - 2)$
$\Rightarrow {a_n} = - 100$
Here ${a_n}$ is the number of notes which cannot be negative so we can say that $n = 24$

So the total time taken by the person to count all the notes is $10 + 24 = 34$ minutes as it takes \[10\] minutes to count first $1500$ notes and $24$ minutes to calculate the last $3000$ notes.

Note:
If we are given the first term and the last term of the AP as $a,a + d,a + 2d,........,l$
Where a is the first term and d is the common difference and l is the last term then we can say that the sum of the AP can be written as
${S_n} = \dfrac{n}{2}(a + l)$