1. 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!

2. 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:

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

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

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

3. Hello, zeromeyzl!

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle ar^2 = b + 25d$ .... (3)
It follows that:

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

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

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

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

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

Therefore ....