Home
Class 12
MATHS
A person is to count 4500 currency notes...

A person is to count 4500 currency notes. Let an denote 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 A.P. with common difference -2, then the time taken by him to count all notes is (1) 34 minutes (2) 125 minutes (3) 135 minutes (4) 24 minutes

Text Solution

AI Generated Solution

To solve the problem step-by-step, we will follow the information given and apply the concepts of arithmetic progression (AP) and summation. ### Step 1: Understand the given information We know that: - The number of notes counted in the first 10 minutes (a1 to a10) is constant and equal to 150. - Therefore, the total number of notes counted in the first 10 minutes is: \[ S_{10} = 10 \times 150 = 1500 ...
Promotional Banner

Topper's Solved these Questions

  • RELATIONS AND FUNCTIONS

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise All Questions|3 Videos

  • STATISTICS

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise All Questions|2 Videos

Similar Questions

Explore conceptually related problems

A person is to count 4500 currency notes. Let a_n , denote the number of notes he counts in the nth minute if a_1=a_2=a_3=..........=a_10=150 and a_10,a_11,......... are in an AP with common difference -2 , then the time taken by him to count all notes is :- (1) 24 minutes 10 11 (2) 34 minutes (3) 125 minutes (4) 135 minutes

A cricketer has to score 4500 runs. Let a _(n) denotes the number of runs he scores in the n ^(th) match. If a _(1)=a_(2)= …. a _(10) =150 and a _(10) , a _(11), a_(12)…. are in A.P. with common difference (-2) . If N be the total number of matches played by him to scoere 4500 runs. Find the sum of the digits of N.

Let the sum of the first n terms of a non-constant A.P., a_(1), a_(2), a_(3),... " be " 50n + (n (n -7))/(2)A , where A is a constant. If d is the common difference of this A.P., then the ordered pair (d, a_(50)) is equal to

If a_1,a_2,a_3,…………..a_n are in A.P. whose common difference is d, show tht sum_2^ntan^-1 d/(1+a_(n-1)a_n)= tan^-1 ((a_n-a_1)/(1+a_na_1))

If a_(1), a_(2), a_(3),........, a_(n) ,... are in A.P. such that a_(4) - a_(7) + a_(10) = m , then the sum of first 13 terms of this A.P., is:

a_(1),a_(2),a_(3),a_(4),a_(5), are first five terms of an A.P. such that a_(1) +a_(3) +a_(5) = -12 and a_(1) .a_(2) . a_(3) =8 . Find the first term and the common difference.

Find the sum of first 24 terms of the A.P. a_1, a_2, a_3, , if it is know that a_1+a_5+a_(10)+a_(15)+a_(20)+a_(24)=225.

If a_1,a_2,a_3,…….a_n are in Arithmetic Progression, whose common difference is an integer such that a_1=1,a_n=300 and n in[15,50] then (S_(n-4),a_(n-4)) is

Find the sum of first 24 terms of the A.P. a_1,a_2, a_3 ......., if it is inown that a_1+a_5+a_(10)+a_(15)+a_(20)+a_(24)=225.

Let n in N . If (1 + x)^n = a_0 + a_1 x + a_2x^2+…. + a_nx^n and a_(n-3), a_(n-2) , a_(n-1) are in A.P then Statement - I : a_1, a_2, a_3 are in A.P. Statement -II : n = 7 The true statements are :