Question

Let 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

• A. AP
• B. GP
• C. HP
• D. none of these

Answer 1

The correct option is D none of these

$x,y,z$ are primes $\Rightarrow \sqrt{x},\sqrt{y},\sqrt{z}$ are irrationals
Let $\sqrt{x}=a+pd;\sqrt{y}=a+qd;\sqrt{z}=a+rd$

Now, $(\sqrt{y}-\sqrt{x}{)}^2=((q-p)d{)}^2$

$\Rightarrow x+y-2\sqrt{xy}=\left(\right(q-p)d{)}^2$

LHS is irrational whereas RHS is rational

Not Possible

Hence, Not in AP
Let $\sqrt{x}=a{r}^p;\sqrt{y}=a{r}^q;\sqrt{z}=a{r}^s$

$\therefore \frac{\sqrt{y}}{\sqrt{x}}=r^{q-p}$

$\Rightarrow y=x\left(r^{2(q-p)}\right)$

$\Rightarrow y$ has two factors $x$ and $r^{2(q-p)}$

Not possible as $y$ is prime

Hence, not in GP
Let $\frac{1}{\sqrt{x}}=a+pd;\frac{1}{\sqrt{y}}=a+qd;\frac{1}{\sqrt{z}}=a+rd$

$\frac{1}{\sqrt{y}}-\frac{1}{\sqrt{x}}=(q-p)d$

$\Rightarrow \frac{(\sqrt{y}-\sqrt{x}{)}^2}{yx}=\left(\right(p-q)d{)}^2$

$x+y-2\sqrt{yx}=xy\left(\right(p-q)d{)}^2$

LHS = irrational and RHS = rational

Not possible. Hence, Not in HP