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(C) Let the three non-zero real numbers in $A.P.$ be $(a - d), a, (a + d)$.Since their squares form a $G.P.$,we have $(a - d)^2, a^2, (a + d)^2$ in $G

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(C) Let the three non-zero real numbers in $A.P.$ be $(a - d), a, (a + d)$.Since their squares form a $G.P.$,we have $(a - d)^2, a^2, (a + d)^2$ in $G.P.$Therefore,$(a^2)^2 = (a - d)^2(a + d)^2$.$a^4 = ((a - d)(a + d))^2 = (a^2 - d^2)^2$.$a^4 = a^4 - 2a^2d^2 + d^4$.$d^4 - 2a^2d^2 = 0$.$d^2(d^2 - 2a^2) = 0$.Since the numbers are non-zero,$d 
eq 0$ is not necessarily true,but if $d=0$,the numbers are $a, a, a$,which form a $G.P.$ with common ratio $r = 1$.If $d^2 = 2a^2$,then $d = \pm \sqrt{2}a$.Case $1$: $d = 0$. The numbers are $a, a, a$. The squares are $a^2, a^2, a^2$. Common ratio $r = 1$.Case $2$: $d = \sqrt{2}a$. The numbers are $a(1-\sqrt{2}), a, a(1+\sqrt{2})$. The squares are $a^2(3-2\sqrt{2}), a^2, a^2(3+2\sqrt{2})$. The common ratio $r = \frac{a^2}{a^2(3-2\sqrt{2})} = 3+2\sqrt{2}$.Case $3$: $d = -\sqrt{2}a$. The numbers are $a(1+\sqrt{2}), a, a(1-\sqrt{2})$. The squares are $a^2(3+2\sqrt{2}), a^2, a^2(3-2\sqrt{2})$. The common ratio $r = \frac{a^2}{a^2(3+2\sqrt{2})} = 3-2\sqrt{2}$.Thus,there are $3$ possible common ratios.

Three non-zero real numbers form an $A.P.$ and the squares of these numbers taken in the same order form a $G.P.$ Then the number of all possible common ratios of the $G.P.$ is

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