The $2^{nd}$, $4^{th}$ and $9^{th}$ terms of an AP The $2^{nd}$, $4^{th}$ and $9^{th}$ terms of an arithmetic progression are in geometric progression. Calculate the common ratio of the geometric progression.
My Attempt:
For A.P
$$T_2=a+d$$
$$T_4=a+3d$$
$$T_9=a+8d$$
Now, 
$$a+d, a+3d, a+8d$$
are in GP.
So,
$$\frac {a+3d}{a+d}=\frac {a+8d}{a+3d}$$.
Now, How should I solve further.?
 A: Hint: 
$$\dfrac AC=\dfrac BD=\dfrac{A-B}{C-D}$$ for $A\ne B$
A: Let $T$ be the 4th term and $d$ be the common difference of the AP. 
The 2nd and 9th terms are $T-2d, T+5d$ respectively. The 2nd, 4th and 9th terms of the AP form a GP:
$$T-2d, T, T+5d$$
The common ratio of the GP,  $r$, is given by
$$r=\frac {\overbrace{\;\;\;T\;\;\;}^A}{\underbrace{T-2d}_{C}}=\frac {\overbrace{T+5d}^{B}}{\underbrace{\;\;\;T\;\;\;}_{D}}=\frac {\overbrace{-5d}^{A-B}}{\underbrace{-2d}_{C-D}}=\color{red}{\frac 52}$$
using componendo and dividendo (correction: subtrahendo!) per hint in @lab's answer.

The following is an illustration of such an AP. 
$$1,\color{blue}4,7,\color{blue}{10},13,16,19,22,\color{blue}{25}$$
A: Express the geometric progression
$$a+d=c,\\a+3d=cr,\\a+8d=cr^2.$$
Subtract pairwise to eliminate $a$,
$$c(r-1)=2d,\\cr(r-1)=5d.$$
Take the ratio to eliminate $d$ ($c$ disappears too),
$$r=\frac52.$$
A: $$\frac {a+3d}{a+d}=\frac {a+8d}{a+3d}$$
Apply comonendo and dividendo,
$$\frac {a+3d+a+d}{a+3d-a-d}=\frac {a+8d+a+3d}{a+8d-a-3d}$$
$$\frac {2a+4d}{2d}=\frac{2a+11d}{5d}$$
$$\frac {2a+4d}{2}=\frac{2a+11d}{5}$$
Cross multiply terms,
$$10a+20d=4a+22d$$
$$6a=2d$$
$$a=\frac 13 d$$
Put value of a in ratio,
$$= \frac{a+3d}{a+d}$$
$$= \frac{\frac 13 d+3d}{\frac13+d}$$
$$= \frac{\frac {10}3d}{\frac 43 d}$$
$$= \frac 52$$
A: 
How should I solve further?

Assume $d \ne 0$, otherwise the sequence is constant. 
Now divide each numerator and denominator of your last equation by $d$ to  get
$$\frac {\frac{a}{d}+3}{\frac{a}{d}+1}=\frac {\frac{a}{d}+8}{\frac{a}{d}+3}$$
and substitate $\frac{a}{d}$ by $r$:
$$\frac {r+3}{r+1}=\frac {r+8}{r+3}$$
This gives $r=\frac{1}{3}$.
So we have 
$$\frac{T_4}{T_2}=\frac {a+3d}{a+d}=\frac {\frac{a}{d}+3}{\frac{a}{d}+1}=\frac {r+3}{r+1}=\frac {\frac{1}{3}+3}{\frac{1}{3}+1}=\frac{5}{2}$$
A: Continuing from where you left off:  You have
$$\frac{a+3d}{a+d} = \frac{a+8d}{a+3d}$$
Cross-multiplying gives
$$(a+3d)^2 = (a+d)(a+8d)$$
$$a^2 + 6ad + 9d^2 = a^2 + 9ad + 8d^2$$
$$d^2-3ad = 0$$
$$d(d-3a) = 0$$
We conclude from this that either $d=0$, or $d=3a$.
In the first case, $d=0$, the arithmetic sequence is constant, and therefore the ratio of the geometric progression is just $1$.   (Note, by the way, that the other solutions presented up to this point seem to have missed that possibility -- which, admittedly, is trivial.)
In the second case, $d=3a$, the $2^{nd}$ term of the arithmetic sequence is $a+3a = 4a$, and $4^{th}$ term is $a + 9a = 10a$, and the $9^{th}$ term is $a + 24a = 25a$.  By inspection $4a, 10a, 25a$ are in geometric progression with common ratio $5/2$.
