Arithmetic and Geometric Series

Question

Given that 4th, 9th and 12th term of an arithmetic series equal to the 5th, 8th and 15th terms of a geometric series, show that the common ratio r of the geometric series satisfies $5r^{10}-8r^3+3=0$

So Let the first term and common difference of the AP be a and d, first term of GP be b. then $a+3d=br^4,a+8d=br^7,a+11d=br^{14}$ but I don't know how that leads to the desired result,,

Answer 1

$3$ times first equation $+$ $5$ times third equation $-$ $8$ times second equation.

$$3(a+3d-br^4) + 5(a+11d-br^{14}) - 8 (a+8d-br^7) = 0 \\ \implies 5br^{14} - 8br^7 + 3br^4 = 0 \\ \implies 5r^{10} - 8r^{3} + 3 = 0$$

Answer 2

Let $b$ is a first term of a geometric series.

Thus, $$\frac{br^7-br^4}{5}=\frac{br^{14}-br^7}{3},$$ which gives what you wish.

Answer 3

Assume that $r\ne1$ and $r\ne0$. We can deal with these cases separately. In this case, $r^8-r^5\ne0$, so $$ \frac{r^{15}-r^8}{r^8-r^5}=\frac{g_{15}-g_8}{g_8-g_5}=\frac{a_{12}-a_9}{a_9-a_4}=\frac35\tag{1} $$ means that $$ 5r^{15}-8r^8+3r^5=0\tag{2} $$ Since we have assumed that $r\ne0$, we can divide by $r^5$ to get $$ 5r^{10}-8r^3+3=0\tag{3} $$ $r=1$ provides a solution, and satisfies $(3)$.

$r=0$ provides a solution, but does not satisfy $(3)$. If we want to encompass all solutions, we could use $$ 5r^{11}-8r^4+3r=0\tag{4} $$