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Arithmetic progression :$a_n=\{a,a+d,a+2d,...,a+nd,...\}$

Geometric progression :$b_n=\{b,bq,bq^2,...,bq^n,...\}$

if : $a_r,a_s,a_t,$ be a Geometric progression($b_n$)

then : What is q?

my way : let :$r>s>t$

$$ a_{r} =a +(r-1)d $$ $$ a_{s} =a +(s-1)d $$ $$ a_{t} =a +(t-1)d $$

$$ a_{s}^{2} =a_{r} a_{t} \Rightarrow a^{2} +2ad(s-1)+ d^{2}(s-1)^{2} =a^{2}+ad(r+t-2)+d^{2}(r-1)(t-1)$$

$$ 2a(s-1)+ d(s-1)^{2} =a(r+t-2)+d(r-1)(t-1)=$$ $$ 2as-ar-at=2sd-dt-dr-d s^{2} +drt \Rightarrow $$ $$a= \frac{d(2s-t-r-s^{2} +rt)}{2s-r-t} $$ $$ \frac{a_{r}}{a_{t}} = \frac{a +(r-1)d}{a +(t-1)d} $$ $$ \frac{d(2s-t-r-s^{2} +rt)}{2s-r-t}+ (r-1)d= $$ $$d \frac{2s-t-r-s^{2} +rt+2rs- r^{2}-rt-2s+r+t }{2s-r-t} = $$ $$ d \frac{-s^{2} -d \frac{ (r-s)^{2}}{2s-r-t} +2rs- r^{2}}{2s-r-t}=-d \frac{ (r-s)^{2}}{2s-r-t} $$

$$ -d \frac{ (s-t)^{2}}{2s-r-t} $$

$$ q^{2}=\frac{a_{r}}{a_{t}}= \frac{(r-s)^{2}}{(s-t)^{2}} $$

There is simpler method?

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    $b_n$ plays no part in this question. q is the common ratio of the target GP. – 2017-02-12
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    The calculation can be slightly shortened if we let R = r - 1, S..., T... – 2017-02-12

2 Answers 2

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We have that $a_r = a + (r-1)d$ and that $a_s = a+(s-1)d$. Thus, $$a_r - a_s = (r-s)d \Rightarrow r-s = \frac{a_r - a_s}{d} = \frac{a_r(1-q)}{d} \tag{1}$$ since, $a_s = a_r (q)$ and $a_t = a_r (q)^2$ where $q$ is the common ratio of the G.P.

Similarly, $\displaystyle s-t = \frac{a_s(1-q)}{d} \tag{2}$ Thus, $$\frac{(1)}{(2)} = \frac{r-s}{s-t} = \frac{a_r}{a_s} = \frac{1}{q}$$ and I hope you can take it from here.

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Let the three terms in the AP be $$\begin{aligned} &a_r=A-(s-r)d\\ &a_s=A\\ &a_t=A+(t-r)d\end{aligned}$$

Common ratio of GP, $$q=\frac {\overbrace{\;\;\;A\;\;\;}^P}{\underbrace{A-(s-r)d}_{R}}=\frac {\overbrace{A+(t-s)d}^Q}{\underbrace{\;\;\;A\;\;\;}_S}=\color{red}{\frac {\overbrace{t-s}^{\color{green}{Q-P}}}{\underbrace{s-r}_{\color{green}{S-R}}}}$$

(see also my solution to another recent question which is quite similar)