Question

Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is
$
  {\text{A}}{\text{. }}\sqrt2 + \sqrt 3 \\
  {\text{B}}{\text{. 3 + }}\sqrt 2 \\
  {\text{C}}{\text{. 2 - }}\sqrt3 \\
  {\text{D}}{\text{. 2 + }}\sqrt 3 \\
 $

Answer 1

Hint: We will proceed by making conditions from the given statement in terms of common ratio and then solve further. It will be easy if we let the three terms in G.P. be $a,ar,a{r^2}$.

Complete step-by-step answer:
Given that if middle term in the given G.P. is doubled then the new numbers will be in A.P.
Let the three terms in G.P. be $a,ar,a{r^2}$
The middle term is \[ar\]
According to the statement given in the question
$ \Rightarrow 2(2ar) = a + a{r^2}$
Simplifying the above equation, we obtain
$
   \Rightarrow 4ar = a(1 + {r^2}) \\
   \Rightarrow 4r = 1 + {r^2} \\
   \Rightarrow {r^2} - 4r + 1 = 0...............................(1) \\
$
The above equation is a quadratic equation in terms of r.
Comparing the equation 1 with \[a{x^2} + bx + c = 0\] ,we get
$a = 1,b = - 4,c = 1$
The solution of a quadratic equation is given by
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Therefore substituting the values of a, b, c in above equation will give the value of r (common ratio)
$
   \Rightarrow r = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4 \times 1 \times 1} }}{{2(1)}} \\
   \Rightarrow r = \dfrac{{4 \pm \sqrt {16 - 4} }}{2} \\
   \Rightarrow r = \dfrac{{4 \pm 2\sqrt3 }}{2} \\
   \Rightarrow r = 2 \pm \sqrt3 \\
   \Rightarrow r = 2 + \sqrt3 \\
$
The value of r is $2 + \sqrt 3 $, the negative value is neglected because the G.P. is increasing and $2 - \sqrt3 < 1$.
Hence the correct option is D.

Note: The general form of G.P. is $a,ar,a{r^2}...............$ where a is the first term and r is the common ratio. To solve some problems in G.P, it is convenient to let G.P. as \[a{r^{ - 1}},a,ar\] , if we multiply these terms the common ratio will be cancelled and only the cube of the first term will be left. This will come in handy where we have to find the first term.