A sequence includes $a_p=\sqrt2$, $a_q=\sqrt3$, $a_r=\sqrt5$ for some $1\leq p
In a sequence $a_1, a_2,\dots$ of real numbers it is observed that $a_p=\sqrt{2}$, $a_q=\sqrt{3}$, and $a_r=\sqrt{5}$, where $1\leq p<q<r$ are positive integers. Then $a_p$, $a_q$, $a_r$ can be terms of
(A) an arithmetic progression
(B) a harmonic progression
(C) an arithmetic progression if and only if $p$, $q$, and $r$ are perfect squares
(D) neither an arithmetic progression nor a harmonic progression
I have tried using the definition of AP and argued that if the first option is true then $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{5}-\sqrt{3}}=\frac{q-p}{r-q}$  is true which implies that the left hand side is an rational number. I don't know whether the left hand side is a rational number or not. I am stuck here. Also, I don't know how correct this approach is. How do I approach and solve this problem?
 A: As mentioned in the comments, by rationalizing the denominator, we can reduce the problem to checking whether $\sqrt{15}-\sqrt{10}-\sqrt{6}$ is rational.  This in turn can be solved as follows:

Set $x = \sqrt{15} - \sqrt{10} - \sqrt{6}$ and suppose $x$ is rational.  Then
$$x^2 = 31 - 10\sqrt{6} - 6\sqrt{10} + 4\sqrt{15} \implies y = \frac{x^2 - 31}{2} = 2\sqrt{15}-3\sqrt{10}-5\sqrt{6}$$ shows that $y$ is rational.  Therefore, $$(3x-y)^2 = (\sqrt{15}+2\sqrt{6})^2 = 39 + 12\sqrt{10}$$ is rational, showing that $$\sqrt{10} = \frac{(3x-y)^2-39}{12}$$ is rational.

I'll leave the irrationality of $\sqrt{10}$ to you (it's the same proof as the irrationality of $\sqrt{2}$).
A: Since $\sqrt{15} - \sqrt{10} - \sqrt{6}$ is a root of $x^4 - 62 x^2 - 240 x - 239$, if it is rational then it must be an integer by the rational root theorem (*). Therefore it's enough to prove that it's not an integer:
$
3.8 < \sqrt{15} < 3.9
$
$
3.1 < \sqrt{10} < 3.2
$
$
2.4 < \sqrt{15} < 2.5
$
$
\implies
-1.9 < \sqrt{15} - \sqrt{10} - \sqrt{6} < -1.6
$
and so $\sqrt{15} - \sqrt{10} - \sqrt{6}$ is not an integer.
Actually, we can avoid these estimates. If $\sqrt{15} - \sqrt{10} - \sqrt{6}$ is an integer, it must divide $239$, which is prime, and so must be $239$. But $239$ is not a root of $x^4 - 62 x^2 - 240 x - 239$.
(*) The precise polynomial is not relevant. What matters is that it is a monic polynomial with integer coefficients.
A: Suppose that $x=\sqrt{15}-\sqrt{10}-\sqrt{6}$ were rational.
Then, squaring the number would give the following:
$$\begin{align}
x^2
&=(\sqrt{15}-\sqrt{10}-\sqrt{6})(\sqrt{15}-\sqrt{10}-\sqrt{6})\\
&=15-\sqrt{150}-\sqrt{90}-\sqrt{150}+10+\sqrt{60}-\sqrt{90}+\sqrt{60}+6\\
&=31-2\sqrt{150}-2\sqrt{90}+2\sqrt{60}\\
&=31-2(5\sqrt{6})-2(3\sqrt{10})+2(2\sqrt{15})\\
&=31-10\sqrt{6}-6\sqrt{10}+4\sqrt{15}
\end{align}$$
Next, to get rid of the $\sqrt{6}$ term, one must multiply $x$ by $-10$ to cancel the $\sqrt{6}$ term from $x^2$. This gives $x^2-10x=(31-10\sqrt{6}-6\sqrt{10}+4\sqrt{15})+(-10\sqrt{15}+10\sqrt{10}+10\sqrt{6})=31+4\sqrt{10}-6\sqrt{15}$.
The above would then lead to the rationality of $31+4\sqrt{10}-6\sqrt{15}$. Since $31$ is clearly rational, we may ignore that term, leading to the rationality of $x^2-10x-31=4\sqrt{10}-6\sqrt{15}$.
Squaring one more time, we get the rationality of $(x^2-10x-31)^2=(4\sqrt{10}-6\sqrt{15})^2=160-48\sqrt{150}+540=700-48(5\sqrt{6})=700-240\sqrt{6}$.
Finally, subtracting $700$ and then dividing by $-240$ leads to the rationality of $\sqrt{6}$.
But we know that square roots of integers are either themselves integers or else irrational, leading to a contradiction. Hence, $x=\sqrt{15}-\sqrt{10}-\sqrt{6}$ must be irrational.
Alternatively, one could instead get rid of the $\sqrt{10}$ term by considering $x^2-6x$, or the $\sqrt{15}$ term by considering $x^2-4x$, and similar steps would show that the original term that one got rid of is rational.
