# Arithmetic and Geometric Series

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,,

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

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

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.

• @Arthur Thank you! I fixed. – Michael Rozenberg Jul 12 '17 at 16:10
• @Arthur Thank you! :) – Michael Rozenberg Jul 12 '17 at 16:14
• Please see comment on @robjohn's solution below. – hypergeometric Jul 12 '17 at 17:00

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 $$5r^{11}-8r^4+3r=0\tag{4}$$

• Thank Youfor the explanation! I have an unanswered question too at math.stackexchange.com/questions/2356355/… – Homaniac Jul 12 '17 at 16:36
• @MichaelRozenberg - does this imply $r\neq 1$? – hypergeometric Jul 12 '17 at 16:36
• @hypergeometric: I have improved the discussion of $r=0$. – robjohn Jul 12 '17 at 16:47
• Noted. But @MichaelRozenberg pointed out in a comment elsewhere (but he is quiet here) that such a solution does not work $r=1$ as it involves division by $0$. – hypergeometric Jul 12 '17 at 16:55
• @hypergeometric: unless Michael Rozenberg has commented on this answer, I don't think paging him here will get to him. – robjohn Jul 12 '17 at 17:04