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!
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: .$\displaystyle a_n \:=\:ar^{n-1} \quad \begin{Bmatrix}a = \text{first term} \\ r = \text{common ratio} \end{Bmatrix}$
For the AP: .$\displaystyle \begin{array}{ccc}a_2&=& a + d \\ a_{10} &=& a + 9d \\ a_{26} &=& a + 25d \end{array}$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.
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]: .$\displaystyle r \;=\;\frac{7d+9d}{7d+d} \:=\:\frac{16d}{8d} \quad\Rightarrow\quad\boxed{ r \:=\:2}$
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 ....