Tricky formula to find common ratio of geometric progression

Question

Suppose $a_1,a_2,a_3,...$ are in an arithmetic sequence ($d\neq 0)$ . If $a_3,a_{13},a_{63}$ are geometric sequence ,find common ratio of geometric sequence .

There is an ordinary method to solve this problem like below $$\quad{\frac{a_{13}}{a_3}=\frac{a_{63}}{a_{13}}=q\\ \frac{a_{13}-a_3}{a_3}=\frac{a_{63}-a_{13}}{a_{13}}\\ \frac{10d}{a_3}=\frac{50d}{a_{13}}\\\frac{10\not d}{a_3}=\frac{50 \not d}{a_{13}}\\ \frac{1}{a_1+2d}=\frac{5}{a_1+12d}\\5a_1+10d=a_1+12d\\4a_1=2d\\2a_1=d \to \\q=\frac{a_{13}}{a_3}=\frac{a_1+12d}{a_1+2d}=\frac{25d}{5d}=5}$$ $Now$ :tricky way $$q=\color{red} {\frac{63-13}{13-3}=\frac{50}{10}=5}$$ My question is : How can prove this tricky method ?

Answer 1

Some hints:$$a_n = a_0 + dn$$

$$a_{63} - a_{13} = (a_0 + 63d) - (a_0 + 13d) = 50d$$

$$\frac{a_{63} - a_{13}}{a_{13} - a_3} = \frac{50d}{10d} = 5$$

Answer 2

There is no trick:

$$ q= {a_{13}\over a_3} = {(q-1)a_{13}\over (q-1)a_3} ={a_{63}-a_{13}\over a_{13}-a_3}={50d\over 10d}$$