0
$\begingroup$

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,,

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @Arthur Thank you! I fixed. $\endgroup$ – Michael Rozenberg Jul 12 '17 at 16:10
  • $\begingroup$ @Arthur Thank you! :) $\endgroup$ – Michael Rozenberg Jul 12 '17 at 16:14
  • $\begingroup$ Please see comment on @robjohn's solution below. $\endgroup$ – hypergeometric Jul 12 '17 at 17:00
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Thank Youfor the explanation! I have an unanswered question too at math.stackexchange.com/questions/2356355/… $\endgroup$ – Homaniac Jul 12 '17 at 16:36
  • $\begingroup$ @MichaelRozenberg - does this imply $r\neq 1$? $\endgroup$ – hypergeometric Jul 12 '17 at 16:36
  • $\begingroup$ @hypergeometric: I have improved the discussion of $r=0$. $\endgroup$ – robjohn Jul 12 '17 at 16:47
  • $\begingroup$ 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 $0$. $\endgroup$ – hypergeometric Jul 12 '17 at 16:55
  • $\begingroup$ @hypergeometric: unless Michael Rozenberg has commented on this answer, I don't think paging him here will get to him. $\endgroup$ – robjohn Jul 12 '17 at 17:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.