Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

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

Option 1)
24 minutes

Option 2)
34 minutes

Option 3)
125 minutes

Option 4)
135 minutes

Answers (1)

best_answer

Sum of n terms of an AP -

$S_{n}= \frac{n}{2}\left [ 2a +\left ( n-1 \right )d\right ]$

and

Sum of n terms of an AP

$S_{n}= \frac{n}{2}\left [ a+l\right ]$

- wherein

$a\rightarrow$ first term

$d\rightarrow$ common difference

$n\rightarrow$ number of terms

General term of an A.P. -

$T_{n}= a+\left ( n-1 \right )d$

- wherein

$a\rightarrow$ First term

$n\rightarrow$ number of term

$d\rightarrow$ common difference

4500=10*150+(148+146+....upto(n-10) terms)

.Let's suppose n minutes taken

$=1500+\frac{\left ( n-10 \right )}{2}\left [2*148+(n-11)(-2) \right ]$

$=1500+(n-10)\left [ 148-n+11 \right ]$

$=1500+159n-1590-n^2+10n$

$\Rightarrow 3000=-n^2+169n-1590$

$\Rightarrow n^2-169n+4590=0$

$\Rightarrow n^2-(135+34)n+4590=0$

$\Rightarrow n(n-34)-135(n-34)=0$

$\Rightarrow (n-34)(n-135)=0$

Now for n=135, $T_{n-10}=T_{125}$

$=148+124*(-2)< 0$ not accepted

Option 1)

24 minutes

Option 2)

34 minutes

Option 3)

125 minutes

Option 4)

135 minutes

Posted by

Himanshu

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2

anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus

anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question