Question 28

# Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is

Solution

Let x = $$a$$, y = $$ar$$ and z = $$ar^2$$
It is given that, 5x, 16y and 12z are in AP.
so, 5x + 12z = 32y
On replacing the values of x, y and z, we get
$$5a + 12ar^2 = 32ar$$
or, $$12r^2 - 32r + 5$$ = 0
On solving, $$r$$ = $$\frac{5}{2}$$ or $$\frac{1}{6}$$

For $$r$$ = $$\frac{1}{6}$$, x < y < z is not satisfied.

So, $$r$$ = $$\frac{5}{2}$$

Hence, option C is the correct answer.

OR