Question

If x, 2y, 3z are in A.P., where the distinct numbers x, y, z are in G.P., then the common ratio of the G.P. is

(a) 3

(b) $\frac{1}{3}$

(c) 2

(d) $\frac{1}{2}$

Answer 1

Let x, 2y, 3z be in A.P, for distinct x, y and z
where x, y, z are in G.P
Since x, 2y and 3z are in A.P
$$\begin{gathered} \therefore 2y=\frac{x+3y}{2}\left(\because \mathrm{middle}\mathrm{term}=\mathrm{arithmetic}\mathrm{mean}\mathrm{of}\mathrm{adjacent}\mathrm{terms}\right) \\ \mathrm{i}.\mathrm{e}4y=x+3y...\left(1\right) \end{gathered}$$
and since x, y and z are in G.P
∴ y² = xz ...(2)
Let r denote the common ratio of G.P
$\therefore \frac{y}{x}=r\mathrm{and}\frac{z}{x}=r^2...\left(3\right)$
Now, dividing (1) by x, we get
$$\begin{gathered} \frac{4y}{x}=1+\frac{3y}{x} \\ \mathrm{i}.\mathrm{e}4r=1+3r^2 \\ \mathrm{i}.\mathrm{e}3r^2-4r+1=0\mathrm{i}.\mathrm{e}3r^2-3r-r+1=0 \\ \mathrm{i}.\mathrm{e}\left(3r-1\right)\left(r-1\right)=0 \\ \mathrm{i}.\mathrm{e}r=\frac{1}{3}\mathrm{or}r=1 \end{gathered}$$
(r = 1 is not possible since x, y, z are distinct)
$\Rightarrow \mathrm{Common}\mathrm{ratio}\mathrm{is}\frac{1}{3}$
Hence, the correct answer is option B.