Analytic way to find geometric progression denominator.

Question

The fourth, sixth, and fourteenth members of a variable arithmetic progression form a geometric progression. Find the denominator of this geometric progression.

What is the most rational way?

Answer 1

We can write the terms of the arithmetic progression like this :

$$a+4b, \ \ a+6b, \ \ a+14b \ \ \ \text{for certain} \ a,b.$$

The constraint is :

$$\dfrac{a+14b}{a+6b}=\dfrac{a+6b}{a+4b}=r \tag{1}$$

which implies

$$\dfrac{a+14b}{a+6b}=\dfrac{a+6b}{a+4b}=\dfrac{8b}{2b}=4=r \ \ \tag{2}$$

Therefore the ratio of the geometric progression is $r=4.$

(2) is due to the following property of fractions :

If $\dfrac{A}{B}=\dfrac{C}{D}$ then $\dfrac{A}{B}=\dfrac{C}{D}=\dfrac{A-C}{B-D}$

Remark : plugging $r=4$ in (1) gives equation

$$3a+10b=0$$

Taking arbitrarily $a=-10$ one deduces $b=3$, which gives the following :

$$a+4b=\color{red}{2}, \ \ \ a+6b=\color{red}{8}, \ \ \ a+14b=\color{red}{32},$$

indeed in geometric progression with ratio $r=4$.

Answer 2

If $a$ is the first term and $d$ is the common difference $$(a+3d)(a+13d)=(a+5d)^2$$

$$\iff16a+39d=10a+25d$$ as $d\ne0$

$$\iff 6a=-14d$$

I believe you need $$\dfrac{a+13d}{a+3d}$$