# APGP Combined

• Nov 1st 2008, 06:47 PM
zeromeyzl
APGP Combined
First 3 terms of a GP are 2nd, 10th and 26th terms of an AP respectively. Show that common ratio of GP is 2.

How the heck do you do that? Thanks for helping! :)
• Nov 1st 2008, 07:18 PM
mr fantastic
Quote:

Originally Posted by zeromeyzl
First 3 terms of a GP are 2nd, 10th and 26th terms of an AP respectively. Show that common ratio of GP is 2.

How the heck do you do that? Thanks for helping! :)

Start by considering the following three equations:

$a = b + d$ .... (1)

$ar = b + 9d$ .... (2)

$ar^2 = b + 25d$ .... (3)
• Nov 1st 2008, 07:32 PM
Soroban
Hello, zeromeyzl!

You're expected to know the basics for an AP and GP.

The $n^{th}$ term of an AP is: . $a_n \:=\:a + (n-1)d\quad\begin{Bmatrix}a = \text{first term} \\ d = \text{common difference}\end{Bmatrix}$

The $n^{th}$ term of a GP is: . $a_n \:=\:ar^{n-1} \quad \begin{Bmatrix}a = \text{first term} \\ r = \text{common ratio} \end{Bmatrix}$

Quote:

The first 3 terms of a GP are 2nd, 10th and 26th terms of an AP respectively.
Show that common ratio of GP is 2.

For the AP: . $\begin{array}{ccc}a_2&=& a + d \\ a_{10} &=& a + 9d \\ a_{26} &=& a + 25d \end{array}$

Since these are in GP: . $\begin{array}{cccccccc}(a+d)r &=& a+9d & \Rightarrow & r &=& \dfrac{a+9d}{a+d} & {color{blue}[1]}\\ \\[-3mm]
(a+9d)r &=& a + 25d & \Rightarrow & r &=& \dfrac{a+25d}{a+9d} & {\color{blue}[2]} \end{array}$

Equate [1] and [2]: . $\frac{a+9d}{a+d} \:=\:\frac{a+25d}{a+9d} \quad\Rightarrow\quad (a+9d)^2 \:=\:(a+d)(a+25d)$

. . $a^2 + 18ad + 81d^2 \:=\:a^2 + 26ad + 25d^2 \quad\Rightarrow\quad 56d^2 \:=\:8ad$

Since $d \ne 0$, we can divide by $8d\!:\;\;7d \:=\:a$

Substitute into [1]: . $r \;=\;\frac{7d+9d}{7d+d} \:=\:\frac{16d}{8d} \quad\Rightarrow\quad\boxed{ r \:=\:2}$

• Nov 1st 2008, 07:35 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
Start by considering the following three equations:

$a = b + d$ .... (1)

$ar = b + 9d$ .... (2)

$ar^2 = b + 25d$ .... (3)

It follows that:

$r = \frac{b + 9d}{b+d}$

$\Rightarrow r(b + d) = b + 9d$

$\Rightarrow rb + rd = b + 9d \Rightarrow b(r-1) = d(9 - r)$

$\Rightarrow \frac{b}{d} = \frac{9-r}{r-1}$ .... (A)

$r = \frac{b + 25d}{b+9d} \Rightarrow \frac{b}{d} = \frac{\text{Left for you}}{r-1}$ .... (B)

Therefore ....