Question

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

Accepted Answer

D. Let $y = xr$ and $z = xr^2$ with $r > 1$ since $x < y < z$. Since $5x, 16y, 12z$ are in AP: $2(16y) = 5x + 12z \Rightarrow 32(xr) = 5x + 12(xr^2)$. Since $x > 0$, divide by $x$: $12r^2 - 32r + 5 = 0$ $12r^2 - 30r - 2r + 5 = 0 \Rightarrow (2r - 5)(6r - 1) = 0$ $r = 5/2$ or $r = 1/6$. Since $r > 1$, $r = 5/2$.

GP and AP Combined Common Ratio

Explanation

Practise this under exam conditions