Question

A person counts $4500$ currency notes . Let ${a_n}$ denote the number of notes the counts in the ${n^{th}}$ minute . If ${a_1} = {a_2} = ............. = {a_{10}} = 150$ and ${a_{10}},{a_{11}},...........$ are in A.P. with common difference $- 2$ , then the time taken by him to counts all notes is
A $34$ min
B $125$ min
C $135$ min
D $24$ min

Answer 1

In this question for the first $10$ min he will count $10 \times 150 = 1500$ currency and then for the remaining currency $3000$ he will count in A.P as $148,146,144.........$ currency in ${a_{11}},{a_{12}},...........$
Use the summation formula for A.P as ${S_n} = \dfrac{n}{2}\left( {2a + (n - 1)d} \right)$ $a = 148,d = - 2,{S_n} = 3000$ .
Hence find n .

Complete step-by-step answer:
As it is given in the question that ${a_n}$ denotes the number of notes the counts in the ${n^{th}}$ minute .
Hence
${a_1} = {a_2} = ............. = {a_{10}} = 150$
So in the First $10$ minutes he will count $10 \times 150 = 1500$ currency
Number of remaining currency is $4500 - 1500$ $= 3000$
Now from the question it is given that ${a_{10}},{a_{11}},...........$ are in A.P with common difference $- 2$
So in ${11^{th}}$ he is count $148$ notes
In ${12^{th}}$ he counts $146$ notes and so on ..........
The series become $148,146,144.........$
So from the summation formula of A.P is ${S_n} = \dfrac{n}{2}\left( {2a + (n - 1)d} \right)$
Here $a = 148,d = - 2,{S_n} = 3000$ n is unknown that we have to find
By putting this value in the equation
$3000 = \dfrac{n}{2}\left( {2 \times 148 + (n - 1)( - 2)} \right)$
$6000 = n\left( {296 + (n - 1)( - 2)} \right)$
$6000 = n\left( {298 - 2n} \right)$
Dividing by $2$ to the whole equation .
$3000 = n\left( {149 - n} \right)$
$3000 = 149n - {n^2}$
${n^2} - 149n + 3000 = 0$
Now for solving the quadratic equation as $24 \times 125 = 3000,125 + 24 = 129$
So ${n^2} - 125n - 24n + 3000 = 0$
$n(n - 125) - 24(n + 125) = 0$
$(n - 125)(n - 24) = 0$
Hence $n = 125,24$
As for $n = 125$ the term ${a_n} = a + (n - 1) \times d$
${a_{125}} = 148 - 147 \times 2$ its value is negative hence $n = 125$min is not possible .
Hence $n = 24$ min for counting the remaining currency .
Total time will be $24 + 10 = 34$ min.

Hence option A will be the correct answer.

Note: If a constant is added or subtracted from each term of an AR then the resulting sequence is an AP with the same common difference.
If each term of an AP is multiplied or divided by a non-zero constant, then the resulting sequence is also an AP.
Always remember As in the last step we got $n = 125,24$ as two values will not be possible . So we have to check the term by the formula ${a_n} = a + (n - 1) \times d$ Hence one will be negative.