A person has 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- a 10-150 and a 10- a 11- a 12- - are in A-P- with common difference as -2- then the time in minutes taken by him to count all notes- isA- 24B- 135C- 34D- 125

The correct option is C 34Till 10^th minute, number of notes counted =1500 ∴ 3000=n2[2(148)+(n 1)( 2)] ⇒ 3000=n(149 n) ⇒ n^2 149n+3000=0 ⇒ (n 125)(n 24)=0 n=12 ...

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Standard XII

Mathematics

Sum of n Terms

A person has ...

Question

A person has 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 $a_{10}, a_{11}, a_{12}, \dots$ are in A.P. with common difference as $- 2$ , then the time (in minutes) taken by him to count all notes, is

24

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34

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125

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135

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Solution

The correct option is B $34$
Till $10^{th}$ minute, number of notes counted $= 1500$
$∴ 3000 = \frac{n}{2} [2 (148) + (n - 1) (- 2)]$
$\Rightarrow 3000 = n (149 - n)$
$\Rightarrow n^{2} - 149 n + 3000 = 0$
$\Rightarrow (n - 125) (n - 24) = 0$
$n = 125$ is not possible as then number of notes counted $< 0$
$\Rightarrow n = 24$
$∴$ Total time required $= 24 + 10 = 34$ minutes

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A person has to count 4500 currency notes. Let an denote the number of notes he counts in the nth minute. If a1=a2=⋯=a10=150 and a10, a11, a12,… are in A.P. with common difference as −2, then the time (in minutes) taken by him to count all notes, is

The correct option is B 34Till 10th minute, number of notes counted =1500 ∴ 3000=n2[2(148)+(n−1)(−2)] ⇒3000=n(149−n) ⇒ n2−149n+3000=0 ⇒ (n−125)(n−24)=0 n=125 is not possible as then number of notes counted <0 ⇒n=24 ∴ Total time required =24+10=34 minutes