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.

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". ]
