<s>[INST] Let $a, b$ and $c$ be the $7$th, $11$th and $13$th terms respectively of an AP. If these are three consecutive terms of a GP, then $\cfrac{a}{c}$ is 
Let $a, b$ and $c$ be the $7$th, $11$th and $13$th terms respectively of a non-constant AP. If these are also the three consecutive terms of a GP, then $\cfrac{a}{c}$ is equal to:

APPROACH:
$a = A+6d  = a\ \ \ $ ; $ \ \ \ b = A+10d = ar \ \ \ $ ; $\ \ \ c = A+12d = ar^2$
Where: $A$ is the first term of the AP, $ \ \ d$ is the common difference of the AP, and $ \ \ r$ is the common ratio of the GP.
$\cfrac {c}{a}= \dfrac {1}{r^2}$
Also,
$$(A+6d)r = (A+10d) \implies Ar+6rd = A+10d $$
$$\implies \cfrac{A}{d} = \cfrac{6r-10}{r-1}$$
Now, we know that:
$$(A+10d)^2 = (A+6d)(A+12d)$$
$$[k(6r-10)+ 10k(r-1)]^2 = [k(6r-10)+6k(r-1)] \ \cdot \ [k(6r-10)+12k(r-1)]$$
where $k$ is the proportionality constant.
$k$ gets cancelled leaving us with the following:
$$(16r - 20)^2 = (12r-16)(18r-22)$$
on solving further, I got:
$$ 5r^2 - 11r +6 = 0$$
Which gives $r = \cfrac{6}{5}$ or  $5$.
so, $\cfrac{1}{r^2} = \cfrac{25}{36}$ or $\cfrac{1}{25}$.
Unfortunately, this isn't the answer. I do not see any flaw in my reasoning. Please help me with my mistake and suggestions of better methods would be most appreciated.
 [/INST] It is unfortunate that the complete question was not posted, though one might suspect this was a multiple-choice question; it would be helpful, but not essential (as we'll see), to know whether we are to arrive at a numerical value or just an expression in terms of the arithmetic difference and/or the geometrical factor.
In any case, it seems likely that we won't actually care what the values of $ \ a \ $ and $ \ c \ $ are.  So saying, we will try to work as far as possible only with the ratio $ \ \frac{a}{c} \ \ . $  It will be helpful to express the specified terms as  $ \ \ c-6d \ , \ c - 2d \ , \ c \ \ . $  We can then describe the geometric ratio  $ \ r \ $ by
$$ \frac{a}{c} \ = \ \frac{1}{r^2} \ \ , \ \ \frac{a}{c-2d} \ = \ \frac{1}{r} $$ $$ \Rightarrow \ \ \frac{a}{c} \ = \ \left( \ \frac{a}{c-2d} \ \right)^2 \  = \ \left(\frac{a}{c} \right)^2 ·   \frac{1}{ \left[ \ 1 \ - \ 2· \left(\frac{d}{c} \right) \ \right]^2}   \ \ \Rightarrow \ \ \frac{a}{c}  \ = \ \left[ \ 2\left(\frac{d}{c} \right) \ - \ 1 \ \right]^2 \ \ . $$
It remains to find a value or expression for $  \left(\frac{d}{c} \right) \ . $  Returning to the ratio between the indicated terms, we have
$$ r \ = \ \frac{c-2d}{c-6d} \ = \ \frac{c}{c-2d} \ \ \Rightarrow \ \ c^2   -   6cd \ = \ c^2 - 4cd + 4d^2 $$ $$ \Rightarrow \ \ 4d \ = \ -2c \ \ \Rightarrow \ \ \frac{d}{c} \ = \ -\frac12 \ \ \Rightarrow \ \ \frac{a}{c}  \ = \ \left[ \ 2\left(-\frac12 \right) \ - \ 1 \ \right]^2  \ = \ \mathbf{4} \ \ . $$
We see from this that the arithmetic series is decreasing , with the selected terms being $$ c-6d \ = \ c \ - \  6·\left(-\frac12 c \right) \ = \ 4c \ \ , \ \  c-2d \ = \ c \ - \  2·\left(-\frac12 c \right) \ = \ 2c \ \ , \ \ c \ \ ,  $$
confirming the geometric progression among these.
[It might be remarked that the issue of a "constant" series was avoided in this development.  We can see, however, that setting $ \ d = 0 \ $ does imply that $ \ \frac{a}{c}  \ = \ \left[ \ 2\left(\frac{0}{c} \right) \ - \ 1 \ \right]^2  \ = \ 1 \ \ , $ a separate indication that this expression "behaves correctly". ]
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