Find three numbers in GP such that their sum is 14 and the sum of their square is 84 –
$\begin{align}
  & \text{A}\text{. }3,6,12 \\
 & \text{B 2,6,18} \\
 & \text{C}\text{. 1,3,9} \\
 & \text{D}\text{. 2,4,8} \\
\end{align}$

Hint: To solve this type of tricky question, first we will take 2 equations by the data given in the question. i.e. $a+ar+a{{r}^{2}}=14$ and ${{a}^{2}}+{{\left( ar \right)}^{2}}+{{\left( a{{r}^{2}} \right)}^{2}}=84$. Then we will square both the sides of equation (1) and simplify it by taking the identity ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca$ and substitute the equation (2) to get the value of ‘a’. Then, with the help of ‘a’ we can easily find the value of ‘r’ by substituting the value of ‘a’, hence we can find the three numbers of GP by substituting both the values in equation (1).

Complete step-by-step answer:
It is given in the question that the sum of the three numbers in GP is 14 and the sum of their square is 84. Let us consider as the first term be ‘a’ and the common ration be ‘r’ then –
$a+ar+a{{r}^{2}}=14$ ………………………… (1)
And ${{a}^{2}}+{{\left( ar \right)}^{2}}+{{\left( a{{r}^{2}} \right)}^{2}}=84$ ……………………….. (2)
By squaring on both sides in equation (1), we get –
\[{{\left( a+ar+a{{r}^{2}} \right)}^{2}}={{\left( 14 \right)}^{2}}\]
We know that –
${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca$
By applying this identity, we get –
$\Rightarrow {{a}^{2}}+{{\left( ar \right)}^{2}}+{{\left( a{{r}^{2}} \right)}^{2}}+2\left( a \right)\left( ar \right)+2\left( ar \right)\left( a{{r}^{2}} \right)+2\left( a{{r}^{2}} \right)\left( a \right)=196$
$\Rightarrow {{a}^{2}}+{{\left( ar \right)}^{2}}+{{\left( a{{r}^{2}} \right)}^{2}}+2{{a}^{2}}r+2{{a}^{2}}{{r}^{3}}+2{{a}^{2}}{{r}^{2}}=196$
By taking ‘2’ as common in the above equation, we get –
$\Rightarrow {{a}^{2}}+{{\left( ar \right)}^{2}}+{{\left( a{{r}^{2}} \right)}^{2}}+2\left( {{a}^{2}}r+{{a}^{2}}{{r}^{3}}+{{a}^{2}}{{r}^{2}} \right)=196$
By substituting equation (2) in the above equation, we get –
$\Rightarrow 84+2\left( {{a}^{2}}r+{{a}^{2}}{{r}^{3}}+{{a}^{2}}{{r}^{2}} \right)=196$
By subtracting both sides by 84, we get –
$\begin{align}
  & \Rightarrow 2\left( {{a}^{2}}r+{{a}^{2}}{{r}^{3}}+{{a}^{2}}{{r}^{2}} \right)=196-84 \\
 & \Rightarrow 2\left( {{a}^{2}}r+{{a}^{2}}{{r}^{3}}+{{a}^{2}}{{r}^{2}} \right)=112 \\
\end{align}$
By dividing both sides by ‘2’ we get –
\[{{a}^{2}}r+{{a}^{2}}{{r}^{3}}+{{a}^{2}}{{r}^{2}}=\dfrac{112}{2}\]
\[{{a}^{2}}r+{{a}^{2}}{{r}^{3}}+{{a}^{2}}{{r}^{2}}=56\]
By taking ‘ar’ common in the above equation, we get –
$ar\left( a+a{{r}^{2}}+ar \right)=56$
By arranging the above equation, we get –
$ar\left( a+ar+a{{r}^{2}} \right)=56$
By substituting equation (1) in the above equation, we get –
$\Rightarrow ar\left( 14 \right)=56$
By dividing both sides by 14, we get –
$\Rightarrow ar=\dfrac{56}{14}$
$\therefore ar=4$ ………………. (3)
Now we will find the values of a and r by taking the equation (1)
$a+ar+a{{r}^{2}}=14$
By substituting the equation (3) in the above equation, we get –
$\Rightarrow a+4+4r=14$
By subtracting 4 from both sides, we get –
$\begin{align}
  & \Rightarrow a+4r=14-4 \\
 & \Rightarrow a+4r=10 \\
\end{align}$
By subtracting 4r from both sides, we get –
$a=10-4r$ …………………. (4)
Now, we will substitute equation (4) in equation (3), so we get –
$\begin{align}
  & \Rightarrow \left( 10-4r \right)r=4 \\
 & 10r-4{{r}^{2}}=4 \\
\end{align}$
By dividing both sides by ‘2’, we get –
$5r-2{{r}^{2}}=2$
By arranging the above equation, we get –
$2{{r}^{2}}-5r+2=0$
By factorizing the quadratic equation, we get –
$\begin{align}
  & \Rightarrow 2{{r}^{2}}-r-4r+2=0 \\
 & \Rightarrow r\left( 2r-1 \right)-2\left( 2r-1 \right)=0 \\
\end{align}$
By taking the common factors, we get –
$\left( 2r-1 \right)=0\text{ or }\left( r-2 \right)=0$
$\begin{align}
  &\Rightarrow 2r-1=0\text{ or }r-2=0 \\
 &\Rightarrow r=\dfrac{1}{2}\text{ or }r=2 \\
\end{align}$
Let us take $r=2$
If $r=2$ then
 $\begin{align}
  & a=10-4r \\
 &\Rightarrow a=10-4\left( 2 \right) \\
 &\Rightarrow a=10-8 \\
 &\Rightarrow a=2 \\
\end{align}$
Then the GP be $a,ar,a{{r}^{2}}=2,4,8$
If $r=\dfrac{1}{2}$ then
 $\begin{align}
  & a=10-4r \\
 &\Rightarrow a=10-4\left( \dfrac{1}{2} \right) \\
 &\Rightarrow a=10-2 \\
 &\Rightarrow a=8 \\
\end{align}$
Then the GP be $a,ar,a{{r}^{2}}=8,4,2$
So, both values of r are possible and it will give us a GP with same terms, but in a different order.
Hence, option D. is the correct answer.

Note: Students can also solve this question by taking the options given in them into consideration. Students can check which option is in the form of a GP. Then they can check the conditions given in the question are justified by which option. For example, it is given in the question that the sum of the three numbers in GP is 14, so we can check the sum of all the options. Such that –
$\begin{align}
  & A.3+6+12=21 \\
 & B.2+6+18=26 \\
 & C.1+3+9=13 \\
 & D.2+4+8=14 \\
\end{align}$
Hence, option D justifies the given question so the option D. is the correct answer.