2000+1000+500+250+125+62.5+.......up to infinite

Sourojit Dhua   10 years AGO

2000+1000+500+250+125+62.5+...
This represents sum of infinite gp where the common divisor is 1/2.

So,sum= a/(1-r) where, a=2000 and r=1/2=0.5

therefore,
sum= 2000/1-0.5 = 4000

Like? Yes (16) | No (2)  

Ravi Bhalani   10 years AGO

2000/(1-0.5)=4000

Like? Yes (1) | No   

Rishabh   10 years AGO

2000+1000+500+....
2000+2000/2+2000/4+...
2000(1+1/2+1/4+...)
2000{1/(1-0.5)}
2000*2
4000
reason:-1+1/2+1/4+1/8...
is an infinite gp series with c.d. 1/2
therefore summation of infinite gp series is a/a-r
i.e. first term/(first term - common difference)

Like? Yes (1) | No   

R. Divya   10 years AGO

first number is 2000 and from that each preceding number is half of the previous number. If you add all those numbers we will be getting 3999.99895. while rounding the number we get 4000 as the answer.

Like? Yes (1) | No   

Sakshi Sao   10 years AGO

Its a gp

Like? Yes (1) | No   

piyush khullar   10 years AGO

sum=a/1-r

Like? Yes (1) | No   

vidhijain   10 years AGO

since it is a infinite gp with r=1/2

Like? Yes (1) | No   

sivaramakrishna   10 years AGO

terms are in G.P.
sum of terms upto infinity in G.P...=a/(1-r)
a=2000
r=1/2
==>2000*2=4000

Like? Yes (1) | No   

shruti   10 years AGO

this is a gp with a common ratio equal to 1/2.so sum of infinite terms of gp is a/(1-r)=2000/(1-0.5)

Like? Yes (1) | No   

suman kumar pan   10 years AGO

formula= a*(1-r^n)/(1-r)
where a= 2000,r=0.5,n=infinite..
so the sum will be =4000...

Like? Yes (1) | No   

Kaushlesh Kumar   10 years AGO

This is a GP series, first term 2000 and common ratio 1/2 this series up to infinite.
sum= first_term/(1-common_ratio)
sum=2000/(1-1/2) =2*2000 = 40000

Like? Yes (1) | No   

sravani   10 years AGO

2000+1000+500+250+125+62.5+.........
2000(1+1/2+1/4+1/8+........)
1/2+1/4+1/8...........=(1/2)/1-1/2=1
2000(1+1)=4000

Like? Yes (1) | No   

vinay pal   10 years AGO

use a/(1-r) in gp sum

Like? Yes (1) | No   

Venkateswarlu   10 years AGO

it is an geometric series
sum of infinite terms = a/1-r
2000/(1-(1/2))=4000

Like? Yes (1) | No   

surbala   10 years AGO

the above is a infinite G.P. with common ratio -1

Like? Yes (1) | No   

kavya   10 years AGO

sum of infinite terms in geometric progression is a/(1-r) where a is first term and r is ratio here a=2000 r 1/2 so sum=4000

Like? Yes (1) | No   

Laxmi    10 years AGO

sum=a/1-r for GP series,a=1st term, r=common ratio
sum=2000/(1-1/2)=4000

Like? Yes (1) | No   

priyanka    10 years AGO

we knw tht gp infinte frmula
s=a/1-r
2000/1-(1/2)=4000

Like? Yes (1) | No   

paaredi muthyam reddy   10 years AGO

terms are in geometric progression of common ratio 1/2
sum of infinity terms=a/(1-r)

2000/(1-1/2)=4000

Like? Yes (2) | No (2)  

RANJAN KUMAR PANDIT   10 years AGO

infinite sum of g.p
= a/(1-r)
= 2000/(1-1/2)
=4000

Like? Yes (2) | No (2)  

Soheli Bhaduri   10 years AGO

The numbers are half of the previous number.So,add them and get the result.
Ans:4000 option(8)

Like? Yes (2) | No (3)  

RagunathKC   10 years AGO

gp order...simply 2000*2 ( when a =2000,r=1/2..upto infinite)
sum =a(1-r^n)/1-r...here a=2000...r=1/2..

Like? Yes (1) | No (2)  

Debashis Dey   10 years AGO

the series forms an infinite geometric progression(GP) with common ratio(r)=1/2 hence its sum will be a/1-r=2000/(1-1/2)=4000 a is the 1st term of the series

Like? Yes (3) | No (4)  

Barun Kumar   10 years AGO

The giver series is in G.P. with common ratio 0.5
sum of infinite terms of gp is given by a/1-r
here a=2000 r=0.5
therfore sum =2000/1-0.5=4000
so option 8

Like? Yes (1) | No (2)  

kanchan   10 years AGO

this is in GP with ratio = 0.5 which is smaller than 1.
so the sum will be = (2000)/(1-0.5)=4000

Like? Yes (1) | No (2)  

MRITUNJAY KUMAR   10 years AGO

su mof series in G.P fir infinite terms=a/1-r.
so, r=1/2.
a=2000.
s=2000/1-0.5 =4000.

Like? Yes (1) | No (2)  

NITIN KUMAR GOYAL   10 years AGO

a/(1-r)=sum
a=2000,r=0.5
sum=4000

Like? Yes (1) | No (2)  

Amar   10 years AGO

Sum of GP = a/(1-r)

2000+1000+500+250+125+62.5+.......up to infinite = 2000/(1-0.5)= 4000

Like? Yes (2) | No (3)  

Dr.Dipin Singh   10 years AGO

2000+1000+500+250+125+62.5+.......up to infinite = 2000/(1-0.5)= 4000

Like? Yes (2) | No (3)  

Kowshik Sarker   10 years AGO

2000+1000+500+250+125+62.5+..=2000(1+1/2+1/4+1/8+..)
so we get
2000(1+(1/2)+1/(2^2)+1/(2^3)+...)=2000* 1/[1-(1/2)]
= 2000* 1/(1/2)=2000*2=4000.

Like? Yes (1) | No (2)  

abhishek srivastava   10 years AGO

sum of Infinite geometric progression=a/(1-r)
here.. series is as follows:-
2000 + 2000*1/2 + 2000*1/4...........
a=2000
r=1/2

Like? Yes (1) | No (2)  

Rohan Kumrawat   10 years AGO

by series formula a/1-r

2000/(1-1/2)

Like? Yes (1) | No (2)  

Sundar Kumar   10 years AGO

Answer: option 8) 4000

Like? Yes (1) | No (2)  

Niket saini   10 years AGO

2000/(1-1/2)=4000
Infinite gp
So formula is a/1-r with common difference 1/2

Like? Yes (1) | No (2)  

REHAN AHMAD   10 years AGO

2000(1 + 1/2 + 1/4 + 1/8.........)
series in g.p
so this is equal to 2000(2)=4000

Like? Yes (1) | No (2)  

Avishek Pattanaik   10 years AGO

It\'s a converging G.P. Series
Use the formula Sum=a/(1-r)
a=1st term=2000
r=common ratio=1000/2000=500/1000=250/500.....=0.5

Answer= 2000/(1-0.5)= 4000 (option 8)

Like? Yes (2) | No (3)  

Dinesh thatti   10 years AGO

sum of G.P. to infinite terms
a/(1-r)=2000/(1-1/2)=4000

Like? Yes (1) | No (2)  

ABHIJEET PANDEY   10 years AGO

Ans.4000
Sum=a(1/1-r)
a=2000,
r=1/2.

Like? Yes (1) | No (2)  

Jigyasa Kumar   10 years AGO

infinite series with common ratio 1/2.formula is a/1-r ; r-- common ratio
so answer is 4000

Like? Yes (2) | No (3)  

sugandhi   10 years AGO

2000+1000+500+250+125+.....infinity it goes on if u add until the 0.(decimal values)u ll get 3999 then adding it wil giv oly 4000.

Like? Yes (1) | No (2)  

Aarush Singh   10 years AGO

infinite series with common ratio 1/2.formula is a/1-r ; r-- common ratio
so answer is 4000

Like? Yes (2) | No (3)  

abhilasha   10 years AGO

infinite series with common ratio 1/2.formula is a/1-r ; r-- common ratio
so answer is 4000

Like? Yes (2) | No (3)  

TUSHAR    10 years AGO

IT IS SUM OF INFINITE G.P. SERIES

Like? Yes (1) | No (2)  

vishnuvardhan   10 years AGO

2000[1+1/2+1/4+1/8+1/16+_ infinite]
2000*[1/(1-1/2)] [formula a/(1-r)]
4000

Like? Yes (1) | No (2)  

charitha   10 years AGO

Given series is in G.P
first term a=2000,
r=1/2
sum of infinite series in G.P is a/(1-r)
so, 2000/(1-0.5)=4000

Like? Yes (1) | No (2)  

Gourango   10 years AGO

sum=x+x/2+x/2^2+x/2^3+.....
=x/(1-1/2)
Here x= 2000
so, sum=2000/(1/2)=4000

Like? Yes (1) | No (2)  

Ramesh Sriram   10 years AGO

terms are in geometric progression of common ratio 1/2
sum of infinity terms=a/(1-r)

2000/(1-1/2)=4000

Like? Yes (2) | No (3)  

Anupam Banerjee   10 years AGO

The series can be written as,
2000*(1 + 1/2 +1/4 + 1/8 +........upto infinity)

= 2000 * (1/(1- 1/2))

= 2000 * 2

=4000;

Like? Yes (1) | No (2)  

Nilesh Dattatraya Veer   10 years AGO

2000+1000+500+---=2000+(2000/2)+(2000/4)+--infinity
2000(1+(1/2)+(1/4)+(1/8)+-----)
but (1/2)+(1/4)+(1/8)+(1/16)----infinity=(1/2)/(1-(1/2))=1
so 2000(1+1)
2000*2=4000

Like? Yes (1) | No (2)  

Sudhanshu   10 years AGO

a = first term
r = common ratio
n = number of terms

Sum of infinite terms = a/(1-r) where r < 1
sum=> 2000/(1-(1/2))
=> 2000/(1/2)
=> 2000*2=4000

Like? Yes (1) | No (2)  

PRIYA PANDEY   10 years AGO

gp of infinite series =a/1-r , 2000/1-(1/2) =2000*2 =4000

Like? Yes (1) | No (2)  

ANANDAKUMAR   10 years AGO

2000+1000+500+250+125+62.5+.......+infinitive
2000(1+1\2+1\4+1\8+1\16+......)
2000(1-1\2)power -1
based on binomial theorem
2000(1\2)power -1
so 2000* 2

=4000 ans,,

Like? Yes (1) | No (2)  

Priya Kumar   10 years AGO

a=2000,r=1/2
sum of the G.P up to infinite terms=a/1-r=2000/1-(1/2)=4000

Like? Yes (1) | No (2)  

Ghanshyam Gulati   10 years AGO

sum of infinite G.P
a/1-r
a=2000=first term
r=0.5=common ratio

Like? Yes (1) | No (2)  

ashutosh   10 years AGO

This series in G.P. with r=1/2.
So sum=2000/1-1/2=4000

Like? Yes (1) | No (2)  

Abhishek   10 years AGO

This is an infinite GP series problem.So,
a/c to question
2000(1+1/2+1/4+1/8+.....upto infinity)
2000*(1/1-1/2)
=2000*2
=4000
hence,8)4000 is correct

Like? Yes (1) | No (2)  

amit baranga   10 years AGO

sum to infinity of GP=[a/(1-r)]=[2000/1-0.5]=4000

Like? Yes (3) | No (5)  

SAHIL DIWAN   10 years AGO

its a converging geometric infinite progression
so sum=a/1-r where a is first term n r is common ratio
so acc to ques sum=2000/(1-1/2)=4000

Like? Yes (1) | No (3)  

Vishal Chugh   10 years AGO

Sum of infinite geometric progression is:

S = a/(1-r)
Here a = 2000
r = 1/2
Therefore,
S = 4000

Like? Yes (4) | No (6)  

dheeraj   10 years AGO

infinite gp = a/(1-r)
= 2000/(1-.5)
=4000

Like? Yes (1) | No (3)  

Vibhanshu Shukla   10 years AGO

sum of infinite GP= a/(1-r)
so answer is 4000

Like? Yes (1) | No (3)  

Ann Mary Jolly   10 years AGO

sum = 2000/(1-1/2)= 4000

Like? Yes (1) | No (3)  

Saraswathy   10 years AGO

a = 2000
ar = 1000
r = 1/2
Sum of infinite terms in GP is a/1-r [since r

Like? Yes (1) | No (3)  

kapil singh   10 years AGO

sum of infinity=a/(1-r)
=2000/(1-.5)
=4000

Like? Yes (1) | No (3)  

Tark Patel   10 years AGO

using the formula s = a/(1-r),
a=2000,r=(1/2)
s = 2000 / (1-(1/2)) = 4000

Like? Yes (1) | No (3)  

Yogesh   10 years AGO

simply use the progression...
nearly after 18th term the value is too small..
so that the value is neglected.
and the sum obtained is 4000...

Like? Yes (1) | No (3)  

Varunraj Reddy   10 years AGO

2000+1000+500+250+125+62.5+31.25+15.625+7.8125+....
=4000
Ans:Option 8)

Like? Yes (1) | No (3)  

Abhinav Khare   10 years AGO

because sum of gp to infinite is a/r-1, where -1

Like? Yes (1) | No (3)