If x,y,z be three positive prime numbers. The progression in which `sqrt(x),sqrt(y),sqrt(z)` can be three terms (not necessarily consecutive) is

Options

A. AP
B. GP
C. HP
D. None of these

Answer

D

Explanation

`:.a,b,c` are positive prime numbers.
Let `sqrt(a),sqrt(b),sqrt(c)` are 3 terms of AP. [not necessarily consecutive ]
Then, `sqrt(a)=A+(p-1)D " " "....(i)"`
`sqrt(b)=A+(q-1)D " " "....(ii)"`
`sqrt(c)=A+(r-1)D " " "....(iii)"`
[A and D be the first term and common difference of AP]
`sqrt(a)-sqrt(b)=(p-q)D " " "....(iv)"`
`sqrt(b)-sqrt(c )=(q-r)D " " "....(v)"`
`sqrt(c)-sqrt(a)=(r-p)D " " "....(vi)"`
On dividing Eq.(v), we get
`(sqrt(a)-sqrt(b))/(sqrt(b)-sqrt(c))=(p-q)/(q-r)" " "....(vii)"`
Since,p,q,r are natural numbers and a,b,c are positive prime numbers, so
Eq. (vii) does not hold.
So, `sqrt(a),sqrt(b),sqrt(c)` cannot be the 3 terms of AP. [not necessarily consecutive ]
Similarly, we can show that `sqrt(a),sqrt(b),sqrt(c)` cannot be any 3 terms of GP andHP. [ not necessarily, consecutive].