If k, k^2 and (2k+1) are in arithmetic progression then k-(1/2k) is?

RAKESH   9 years AGO

k^2 - k = 2k + 1 - k^2
=> 2k^2 - 1 = 3k
divide by 2k
=> k - 1/2k = 3/2

Like? Yes (8) | No (4)  

darshana kumrawat   9 years AGO

k^2 - k=2k+1-(k^2)(on solving quadratic eq )
k=(3+(17^1/2))/4 or k=(3-(17^1/2))/4

Like? Yes (2) | No   

KALYAN KUMAR REDDY VALLURU   9 years AGO

As k, k^2 and (2k+1) are in Arithmetic progression, then
(k^2)-k=(2k+1)-(k^2)
=>2(k^2)-1=3k
=>(2(k^2)-1)/3k=2/3

Like? Yes (2) | No (1)  

Kaju Gupta   9 years AGO

k-(1/2k) = (2k^2 - 1)/2k
k, k^2 and (2k+1) are in arithmetic progression then
k^2 - k = 2k+1-k^2
solve this equation , we get
2k^2 - 1 = 3k
divide both side by 2k,
(2k^2 - 1)/2k = 3k/2k = 3/2

option 3)

Like? Yes (2) | No (1)  

aaditya raj   9 years AGO

k^2=(2k+1+k)/2
K^2=(3k+1)/2
now k-1/2k=(2k^2-1)/2 put value of k^2 here and get 3/2

Like? Yes (3) | No (2)  

nitin kishore   9 years AGO

(k^2)-k = 2k+1 - (k^2)
2*(k^2)-3*k-1 =0;
k= 1.78, -.28

using these values,
we get the value og the given expression as 1.5

Like? Yes (1) | No   

praveen kumar   9 years AGO

as the series is in AP
so , k^2-k = 2k+1-k^2
k=1.78

so k-(1/2k) = 1.5

Like? Yes (1) | No   

Pankaj kumar   9 years AGO

A.M.= ((k)+(2k+1))/2 = k^2
so, 2k^2=3k+1
also 2k^2-1=3k
divide both side by 2k
hence (2k^2-1)/(2k)=3k/2k=3/2
and LHS can be written as
k-(1/2k)=3/2.

Like? Yes (1) | No   

sompon temtheerakij   9 years AGO

f k, k^2 and (2k+1) are in arithmetic progression
k^2 - k = (2k+1) - k^2
2k^2 - 3k - 1 = 0
solve this equation --> k = 1.78
k-(1/2k) = 1.78 - 1/(2*1.78)
k = 3/2

Like? Yes (1) | No   

SUNIL GUPTA   9 years AGO

K^2-K=(2K+1)-K^2
2K^2-1=3K
So
k-(1/2k)=(2k^2-1)/2k=3k/2k=3/2

Like? Yes (3) | No (2)  

Yathiraj   9 years AGO

From the given airthmatic progression,
k^2 - k = (2k+1) - k^2 (as the difference should be same in the sequence).
Hence, from the above, we arrive at the quadratic equation
2k^2-3K-1=0
Applying the quadratic equation formula, we get k = (3 +/- sqrt(17)) / 4
and Hence k-(1/2k) = 3/2

Option 3.

Like? Yes (3) | No (2)  

Pranati Kundu   9 years AGO

k^2-k=(2k+1)-k^2
k=1.78,k=-0.28
putting k=1.78
k-(1/2k)=1.5

Like? Yes (1) | No   

arush gupta   9 years AGO

Ans 3/2

Like? Yes (1) | No   

Sanu Panji   9 years AGO

k+2k+1=2k^2
2k^2-1=3k

so
k-(1/2k)
=(2k^2-1)/2k
= 3k/2k
=3/2

Like? Yes (2) | No (2)  

Supriya Mondal   9 years AGO

From the question we can write,
difference between 2nd and 1st term=difference between 3rd and 2nd term
So, k^2-k=2k+1-k^2
so,2k^2-1=3k

We have to find value of (k-1/2k)=(2k^2-1)/2k
=3k/2k [As we get, 2k^2-1=3k]
=3/2
So, option 3)3/2 is the right answer

Like? Yes (5) | No (5)  

sunilbaba    9 years AGO

k^2-k=2k+1-k^2,

Like? Yes (2) | No (2)  

J.DHANDAPANI   9 years AGO

If k, k^2 and (2k+1) are in arithmetic progression,then 2k^2 = 3k + 1
k- (1/2k) = 2k^2-1/2k
= 3k+1-1/2k
=3k/2k = 3/2

Like? Yes (2) | No (2)  

mustain ahmed   9 years AGO

if a,b,c in A.P then b-a=c-b
K^2-K=2k+1-k^2
..........................
(K-1/2k)=3/2.
Hence option-3 is correct answer

Like? Yes (3) | No (3)  

Manish Srivastava   9 years AGO

k=(2k^2-1)/2k;
put in k-i/2k;
3/2

Like? Yes (1) | No (1)  

Dorababu M   9 years AGO

from ap series (2k+1+k)/2=k^2
2k^2-3K-1=0
2k(k-3/2-1/2k)=0
k-3/2-1/2k=0
k-1/2k=3/2
so Option 3 holds the answer

Like? Yes (2) | No (2)  

PAWAN KUMAR   9 years AGO

Since it is an A.P.
so, k^2 - k = 2k+1 - k^2
On solving, we get (2k^2 - 1) = 3k
So,
(2k^2-1)/2k = 3k/2k = 3/2. Ans.

Like? Yes (1) | No (1)  

Ravi P   9 years AGO

2k^2---3k=0
2k(k-1/2k-3/2)=0
k-1/2k=3/2.

Like? Yes (1) | No (1)  

CHAYAN KUMAR   9 years AGO

the no r in ap so,
k^2-k=2*k+1-k^2
solving,k=(3+17^(1/2))/4,(3-17^(1/2))/4
putting this we get 3/2

Like? Yes (1) | No (1)  

arpit shrivastava   9 years AGO

Solve frst eqn and make it equivalent as second eqn. .
we will easily get 3/2

Like? Yes (1) | No (1)  

ganesh babu   9 years AGO

k, k^2, 2k+1 are in AP
k+2k+1=2(k^2) (1st term+ 3rd term= 2(2nd term))
3k+1=2K^2
3k=2k^2-1
3/2(2k)=2k^2-1
3/2=k-(1/2k)

Like? Yes (1) | No (1)  

rayvadera meet    9 years AGO

k+2k+1=2k^2
Therefore
2K^2-1=3k

k-1/2k=k-1/2k=3/2

Like? Yes (2) | No (2)  

Srinivas   9 years AGO

a,b,c are in a.p means b=(a+c)/2
k^2= (k+2k+1)/2
k-(1/2K)=3/2

Like? Yes (1) | No (1)  

Akhil Malhotra   9 years AGO

We have to find k-(1/2k)
=(2k^2-1)/2k

As it is in AP, so
K^2-k=2k+1-k^2
2k^2-1=3k
Dividing both by 2k
We get (2k^2-1)/2k=3k/2k
Hence the answer is 3/2

Like? Yes (1) | No (1)  

SANTOSH SAHOO   9 years AGO

given AP so
k^2-k=(2k+1)-k^2
2K^2-3k-1=0

now we hv to find k-(1/2k)
=(2k^2-1)/2k
=(2k^2-3k-1+3k)/2k
=(0+3k)/2k (as 2k^2-3k-1=0)
=3/2

Like? Yes (1) | No (1)  

Abhik Das   9 years AGO

simply as they are in A.P
so, (3*k+1)=2*k^2
or (2*k^2-1)=3*k
Now for finding {k-(1/2k)}=(2*k^2-1)/2k=3/2

Like? Yes (1) | No (1)  

shobhit ratna mathur   9 years AGO

its too easy

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Jaspalsingh Parmar   9 years AGO

option 2 is correct

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sandeep ramchandra pawaskar   9 years AGO

Option 3 is correct

Like? Yes (1) | No (1)  

Jay parmar   9 years AGO

option 1 is correct

Like? Yes (1) | No (1)  

Surit Datta   9 years AGO

According to the problem
As it is in A.P so,
k^2 - k = (2*k + 1) - k^2
solving this we get,
2*k^2 - 1 = 3*k
we have to find (k - 1/2*k)
which is (2*k^2 - 1)/2*k
as we know (2*k^2 - 1) = 3*k so,
it is 3*k/2*k = 3/2
so option 3
that's it.

Like? Yes (3) | No (4)  

MD SOHAIL AKHTER   9 years AGO

since it is in ap
therefore, k^2-k=2k+1-k^2
=> 2k^2-1=3k
A/Q
2k-1/2k=(2k^2-1)/2k=3/2

Like? Yes (1) | No (2)  

Harish Reddy Namala   9 years AGO

k^2 - k = 2k +1 -k^2
By solving we get k = (3+ (17^0.5)) /4 and
k = (3- (17^0.5)) /4
By substituting either of the k values we get k-(1/2k) = 3/2

Like? Yes (1) | No (2)  

jalaj   9 years AGO

3/2

Like? Yes (1) | No (2)  

Gobind Kumar   9 years AGO

Comman differance should be same so

K^2-k=2K+1-k^2

BY SOLVING IT WE GET
K-(1/2k)=3/2

Like? Yes (1) | No (3)  

nikhil   9 years AGO

[2k(2) - 1]/k = 3

Like? Yes (1) | No (3)  

ABHINAV PUSHPAM   9 years AGO

Answer is 3/2

Like? Yes (1) | No (3)