The sequence k, k^2, 16, k^3 has 4 terms in which first 3 terms are in GP. What should be added in f

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04 Jun 2015 GP and AP

The sequence k, k^2, 16, k^3 has 4 terms in which first 3 terms are in GP. What should be added in fourth term to get the last 3 terms in AP.

OPtion

1) 16
2) 16 - 16^(2/3)
3) 16 - 16^(1/3)
4) 16 + 16^(2/3)
5) 16 + 16^(1/3)
6) 8
7) 8 - 16^(1/3)
8) 8 + 16^(2/3)
9) 8 + 16^(1/3)
10) 8 - 16^(2/3)

Solution

First 3 terms are in GP so (k^2)^2 = k*16
=> k^3 = 16
=> k = 16^1/3
For getting last 3 terms in AP, let x should be added in fourth term
so (k^2 + k^3 + x)/2 = 16
=> (k^2 + k^3 + x) = 32
=> 16^(2/3) + 16 + x = 32
=> x = 16 - 16^(2/3)

Correct Option: 2 Best Solution (20)

Supriya Mondal   9 years AGO

From the question we can write
k^4=k*16
so, k^3=16
so,k=16^(1/3)
Again we can write,
2*16=k^2+(16+x) [Let we have to add x with k^3 and K^3=16]
so, x=16-k^2
=16-16^(2/3)
So, option 2)16-16^(2/3) will be right answer

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Ramdas Singathiya   9 years AGO

k, k^2, 16 are in GP so k^2= sqrt(16*k)
=> k= (16)^1/3

now if k^2,16,k^3 are in ap then 16= (k^2 +k^3)/2
put value of k here
we get that we have to add in right side something
that something will be 16 - 16^(2/3)

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Aashish gohel   9 years AGO

k,k^2,16,k^3
first 3 terms are in GP...
So,
(k^2)^2=16*k
k^4=16*k
k^3=16
k=16^(1/3)......(1)
now if last 3 terms are in AP then we have..
k^2+k^3+x=2*16
where x is the term to be added...
but as per result (1) putting k^3=16,k^2=16^(2/3)...
16^(2/3)+16+x=32
solving it
x=16-16^(2/3)

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W J Krishna Teja   9 years AGO

since first three terms are in g.p , therefore 16*k=(k^2)^(1/2)
that gives k=2*2^(1/3)
so last three terms are 4*2^(2/3) , 16 , 16
let x be added to last term for the three terms to be in a.p.
then x=16-4*2^(2/3)=16-16^(2/3)

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kumari jyoti   9 years AGO

for the 1st 3 terms to be in gp,
k^4=16k
=>k^3=16
for the last 3 terms to be in ap,
16-k^2=k^3+x-16
=>x=16-16^2/3

Like? Yes (1) | No   

Math Lovers   9 years AGO

The sequence k, k^2, 16, k^3 has 4 terms in which first 3 terms are in GP.
It means k, k^2, 16 are in GP.
so (k^2)^2 = 16k
=> k = 16^(1/3)

Now last 3 term k^2, 16, k^3+x terms are in AP.
so 32 = k^2 + k^3 +x

put k =16^(1/3) and solve

x = 16 - 16^(2/3)

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neha gupta   9 years AGO

as first three terms are in GP
therefore:

K^2/k=16/(k^2)

k^4=16K

k^3=16

k=(16)^(1/3)
Now last three terms are to be made in AP by adding something to fourth term
Let the added quantity be x , then
16-(k^2)=(k^3)-16+x
16-(k^2)=16-16+x
16-(k^2)=x
or
x=16-(16^(1/3)^2
x=16-(16^(2/3))

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Dinni   9 years AGO

In g.p, we get k value i. e., k=16^ (1/3).
In a.p, Since common difference is 16-16^ (2/3).
So the 4th term would be 32-16^ (2/3).
Already we have 16 in 4th term by putting k=16^(1/3).
So to get desired value we have to add 16-16^(2/3).

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Jyothi C Allapur   9 years AGO

k , k^2, 16 are in GP
so common ratio is k^2 / k = 16 / k^2
k^3 = 16 that is k = (16) ^ 1/3

k = (16) ^ 1/3 , k ^ 2 = (16) ^ 2/3 , k ^ 3 = 16

and

given that (16) ^ 2/3 , 16 , 16 + x are in AP

so, common difference is given by d = 16 - (16) ^ 2/3 is same as (16 + x ) - 16

i.e. , (16+x) - 16 = 16 - (16) ^ 2/3
x = 16 - (16) ^ 2/3

so , the value to be added to 4th term is 16 - (16) ^ 2/3

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viswambher   9 years AGO

let the first 3 terms be a,ar,ar^2
equating them to k, k^2, k^3
we get r=a=k= 16^(1/3)

and let the last three terms be equal to
p, p+d , p+2d and equating them to k^2, 16 , k^3 +(x)

where k=16^(1/3)
from this we get p=k^2=16^(2/3)
and d =16- 16^(2/3)

k^3+x= p+2d
x=16-16^(2/3)

therefore x must be added to make the last three terms an arithmetic progression

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Mehul D Parmar   9 years AGO

k,k^2,16 in GP

k^2,16.k^3 in AP

then for GP k^2/k=16/k^2

k^3=16.................(1)

now for AP

assume X is requried term

x+k^3=16-K^3

x=16-K^2...........(2)




hence equ (1) and (2)..

required term is

16-16^2/3

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syed kashif ullah   9 years AGO

k^3=16 from gp...and ap common difference =x=16-16^2/3 which is same as no. to be added for making 4th term in ap

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Narelle Morris   9 years AGO

ratio of GP is 16^(2/3), so GP is (16^(2/3), 16, 16. Third term should be 32 - 16^(2/3) so add 16 - 16^(2/3) to the 16 already there.

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manikanta mitta   9 years AGO

three terms are G.P means k^4=16*k => k= 16^(1/3)
then terms become 16^(1/3) , 16^(2/3) ,16 ,16.

To get last term into A.P add common difference d=16 - 16^(2/3)

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ammar   9 years AGO

k^3=16
(k^2+k^3+x)/2=16
or x=16-16^2/3

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ravi singh   9 years AGO

answer is option 2

first 3 are in GP hence

K^3 = 16
and Let x added, since last 3 are in AP hence,
16 - K^2 = (K^3 + x) - 16

then, x = 16 - 16^(2/3)

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akas   9 years AGO

Solu:
K,k^2,16,k^3
First 3 terms are inG.P
K^2/k=16/k^2
K^3=16
K=(16)^1/3
K^2,16,K^3+w are in A.P
16-K^2=k^3+w-16
16-16^2/3=w

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PRASHANT KUMAR   9 years AGO

K=16^(1/3);
SO
LAST TERM TO BE ADDED=16-K^2

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Karthick Raja   9 years AGO

First three terms are in GP so
a=k, ar =k^2, a(r^2)=16
a=k & ar=k^2 sub a=k then r=k
a(r^2)=16
sub a,r value
k*(k^2)=16
k^3=16
k=2.5198420997897
second term
ar=2.5198420997897*2.5198420997897=6.3496042078725

the last three terms are in ap
6.3496042078725 , 16 , (k^3 + x)
k^3=16
6.3496042078725 , 16 , (16 + x)
we can find x
first we can find difference
16-6.3496042078725=9.6503957921274

now 2 nd term will be add with 9.650395792127 so we can check the options which is coming 9.650395792127
take 2 nd option
16 - 16^(2/3) =16- 6.3496042078728=9.650395792127
so the option is 2.

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NITESH DAS   9 years AGO

as it is in gp series,
k=16/k^2
k=16^(1/3)

considering ap series for last 3 term
16*2=k^2+k^3+x
x=16 - 16^(2/3)

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