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Standard X

Mathematics

Arithmetic Progression

A person is t...

Question

A person is to count 4500 currency notes. Let $a_{n}$ denote the number of notes he counts in the $n^{t h}$ minute. If $a_{1} = a_{2} = . . . = a_{10} = 150 a n d a_{10}, a_{11}, . . .$ are in an AP with common difference - 2, then the time taken by him to count all notes is

34 minutes

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125 minutes

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135 minutes

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24 minutes

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Solution

The correct option is A 34 minutes
Till $10^{t h}$ minute number of counted notes = 1500
$3000 = \frac{n}{2} [2 \times 148 + (n - 1) (- 2)] = n [148 - n + 1] n^{2} - 149 n + 3000 = 0 \Rightarrow n = 125, 24$
But n = 125 is not possible
$∴$ total time = 24 + 10 = 34 minutes.

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A person is 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,... are in an AP with common difference - 2, then the time taken by him to count all notes is

The correct option is A 34 minutesTill 10th minute number of counted notes = 1500 3000=n2[2×148+(n−1)(−2)]=n[148−n+1]n2−149n+3000=0⇒n=125,24 But n = 125 is not possible ∴ total time = 24 + 10 = 34 minutes.

A person is to count 4500 currency notes. Let $a_{n}$ denote the number of notes he counts in the $n^{t h}$ minute. If $a_{1} = a_{2} = . . . = a_{10} = 150 a n d a_{10}, a_{11}, . . .$ are in an AP with common difference - 2, then the time taken by him to count all notes is

The correct option is A 34 minutes
Till $10^{t h}$ minute number of counted notes = 1500
$3000 = \frac{n}{2} [2 \times 148 + (n - 1) (- 2)] = n [148 - n + 1] n^{2} - 149 n + 3000 = 0 \Rightarrow n = 125, 24$
But n = 125 is not possible
$∴$ total time = 24 + 10 = 34 minutes.