Three numbers, the third of which 12 form a decreasing G. P. If the last term were 9 instead of 12, they would have formed an A. P. Find the common ratio of the series?
(a) $\dfrac{1}{3}$,
(b) $\dfrac{2}{3}$,
(c) $\dfrac{3}{4}$,
(d) $\dfrac{4}{5}$,

Hint: We start solving the problem by recalling the general form of G.P series. We then find the common ratio in terms of the first term to get out the first condition. We then replace 12 in the series with 9 and use the definition of common difference. We substitute the condition of common difference in this and find the value of the first term. Using the value of the first term, we find the required value of common difference.

Complete step-by-step answer:
According to the problem we have three numbers and if the third number is 12, they form a decreasing G. P. If the last term is replaced with 9, then the numbers form an A. P. We need to find the common ratio of the series.
We know that the general terms of the G. P is defined as $a$, $ar$, $a{{r}^{2}}$, ……, where a is first term and r is the common ratio.
So, let us assume the three numbers are $a$, $ar$ and 12. We know that in a decreasing G. P first term will be greater than the other terms. So, we get $a>12$ ---(1).
We know that the common difference in a G. P is defined as the ratio of the two consecutive terms in G. P.
So, we get $r=\dfrac{12}{ar}$.
$\Rightarrow {{r}^{2}}=\dfrac{12}{a}$.
$\Rightarrow r=\sqrt{\dfrac{12}{a}}$ ---(2).
Now, we replace 12 with 9 in the series to get the required A. P.
So, we have $a$, $ar$, 9 which are in A.P.
We know that the difference between second and first terms in an A.P is equal to the difference between the third and second terms in that A.P which is called a common difference.
So, we get $ar-a=9-ar$.
$\Rightarrow ar+ar=9+a$.
$\Rightarrow 2ar=9+a$.
From equation (1), we have
$\Rightarrow 2a\left( \sqrt{\dfrac{12}{a}} \right)=9+a$.
$\Rightarrow 2\sqrt{12a}=9+a$.
Let us square on both sides.
$\Rightarrow {{\left( 2\sqrt{12a} \right)}^{2}}={{\left( 9+a \right)}^{2}}$.
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
$\Rightarrow 48a=81+{{a}^{2}}+18a$.
$\Rightarrow {{a}^{2}}+18a-48a+81=0$.
$\Rightarrow {{a}^{2}}-30a+81=0$.
$\Rightarrow {{a}^{2}}-27a-3a+81=0$.
$\Rightarrow a\left( a-27 \right)-3\left( a-27 \right)=0$.
$\Rightarrow \left( a-3 \right)\left( a-27 \right)=0$.
$\Rightarrow \left( a-3 \right)=0$ or $\left( a-27 \right)=0$.
$\Rightarrow a=3$ or $a=27$.
From equation (1), we have $a>12$. So, the value of a is 27. We substitute this in equation (2).
$\Rightarrow r=\sqrt{\dfrac{12}{27}}$.
$\Rightarrow r=\sqrt{\dfrac{4}{9}}$.
$\Rightarrow r=\dfrac{2}{3}$.
We have found the common ratio of the series as $\dfrac{2}{3}$.

So, the correct answer is “Option b”.

Note: We need to make sure that the values will be decreasing in a decreasing G.P. We can extend the series after finding the absolute value of the common ratio of the series. Since, the given series is a decreasing G.P we can get a finite value for the sum of infinite terms in G.P. We can also find the 2nd term in the series using the value of common ratio. Similarly, we can expect problems to find the sum of n-terms in A.P, G.P.