If the 5th, 9th, and 16th terms of an arithmetic progression are also consecutive terms of a geometric progression, find the common ratio of the geometric progression.

If $A_5$, $A_9$, and $A_{16}$ are consecutive terms of a GP, then the square of the middle term equals the product of the first and third terms.

$$(A_9)^2 = A_5 \times A_{16}$$

Substitute the expressions for $A_5$, $A_9$, and $A_{16}$:

$$(a + 8d)^2 = (a + 4d)(a + 15d)$$

$$a^2 + 2(a)(8d) + (8d)^2 = a(a + 15d) + 4d(a + 15d)$$ $$a^2 + 16ad + 64d^2 = a^2 + 15ad + 4ad + 60d^2$$ $$a^2 + 16ad + 64d^2 = a^2 + 19ad + 60d^2$$

Subtract $a^2$ from both sides:

$$16ad + 64d^2 = 19ad + 60d^2$$

Rearrange the terms to one side:

$$64d^2 - 60d^2 = 19ad - 16ad$$ $$4d^2 = 3ad$$

Since $d$ cannot be zero (otherwise all terms of the AP would be the same, and thus the GP terms would also be the same, leading to a common ratio of 1, which is a trivial case and usually not what's intended in such problems), we can divide both sides by $d$:

$$4d = 3a$$

Express $a$ in terms of $d$:

$$a = \frac{4d}{3}$$

The common ratio $r$ of the GP is given by $\frac{A_9}{A_5}$.

$$r = \frac{a + 8d}{a + 4d}$$

Substitute $a = \frac{4d}{3}$ into the expression for $r$:

$$r = \frac{\frac{4d}{3} + 8d}{\frac{4d}{3} + 4d}$$

To simplify, find a common denominator for the terms in the numerator and denominator:

$$r = \frac{\frac{4d}{3} + \frac{24d}{3}}{\frac{4d}{3} + \frac{12d}{3}}$$ $$r = \frac{\frac{4d + 24d}{3}}{\frac{4d + 12d}{3}}$$ $$r = \frac{\frac{28d}{3}}{\frac{16d}{3}}$$

Multiply by the reciprocal of the denominator:

$$r = \frac{28d}{3} \times \frac{3}{16d}$$

Cancel out 3 and $d$:

$$r = \frac{28}{16}$$

Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 4:

$$r = \frac{28 \div 4}{16 \div 4}$$ $$r = \frac{7}{4}$$

The common ratio of the GP is $\frac{7}{4}$.