You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

125 minutes

No worries! We‘ve got your back. Try BYJU‘S free classes today!

135 minutes

No worries! We‘ve got your back. Try BYJU‘S free classes today!

24 minutes

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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.

Suggest Corrections

Similar questions

Q. A person is to count

4500

currency notes. Let

a_{n}

denotes 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 common difference

- 2

, then the time taken by him to count all notes is?

Q. 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

Q. 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

Q. A person is to count

4500

currency notes. Let

a_{n}

denote the number of notes he counts in the

n^{t h}

minute. lf

a_{1} = a_{2} = \dots \dots = a_{10} = 150

and

a_{10}, a_{11}, \dots \dots

are in A.P. with common difference

(- 2)

, then the time taken by him to count all notes is

Q. A person is to count

4500

notes. Let

a_{n}

denote the number of notes he counts in the

n^{t h}

minute. If

a_{1} = a_{2} = a_{3} = . . . = a_{10} = 150

, and

a_{10}, a_{11}, a_{12}

,... are in

A P

with the common difference

- 2

, then the time taken by him to count all the notes is

Join BYJU'S Learning Program

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.