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

Question

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.

Answer 1

The value of $\frac Ad$ you have obtained is incorrect. $$\frac Ad=\frac {10-6r}{r-1}$$ is the correct value. Hence the following work becomes incorrect too. $$[k(10-6r)+10k(r-1)]^2=[k(10-6r)+6k(r-1)][k(10-6r)+12k(r-1)]$$ So that $$ 16r^2=4(6r-2)$$ $$ 2r^2=3r-1$$ $$ 2r^2-3r+1=0$$ $$ (2r-1)(r-1)=0$$ So either $\frac ac=1$, or $4$, but $r=1$ will not be considered, as AP is non-constant.

Answer 2

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". ]