Question

If A is the arithmetic mean and $ {G_1},{G_2} $ be two geometric means between any two numbers, then $ \dfrac{{{G_1}^2}}{{{G_2}}} + \dfrac{{{G_2}^2}}{{{G_1}}} $ is equal to:
A. 2A
B. A
C. 3A
D. None of these

Answer 1

Hint: Given A is the arithmetic mean of two numbers, say and b. Find the arithmetic mean of a and b. $ {G_1},{G_2} $ are the geometric means between a and b then a, $ {G_1},{G_2} $ , b are in Geometric progression. Let ‘r’ be the common ratio and find the values of $ {G_1},{G_2} $ in terms of a, b, r and then substitute $ {G_1},{G_2} $ values in $ \dfrac{{{G_1}^2}}{{{G_2}}} + \dfrac{{{G_2}^2}}{{{G_1}}} $ and replace arithmetic mean of a and b with ‘A’ to get the answer.

Complete step-by-step answer:
Let us assume a and b are the two numbers and ‘r be the common ratio.
We are given that A is the arithmetic mean of and b which means $ A = \dfrac{{a + b}}{2} \to 2A = a + b \to eq(1) $
 $ {G_1},{G_2} $ are two geometric means between a and b.
 $ a,{G_1},{G_2},b $ are in Geometric progression.
‘a’ if the first term and ‘b’ is the fourth term in the G.P. and nth term of a geometric progression is
 $ {T_n} = a{r^{n - 1}} $
‘b’ is the fourth term i.e. n=4
 $
  {T_4} = a{r^{4 - 1}} \\
  {T_4} = b \\
  b = a{r^3} \\
  {r^3} = \dfrac{b}{a} \\
  r = \sqrt[3]{{\dfrac{b}{a}}} \\
  $
 $ {G_1} $ is the second term i.e. n=2
 $
  {T_2} = a{r^{2 - 1}} \\
  {T_2} = {G_1} \\
  {G_1} = ar \\
  r = \sqrt[3]{{\dfrac{b}{a}}} \\
  {G_1} = a \times \sqrt[3]{{\dfrac{b}{a}}} \\
  $
 $ {G_2} $ is the third term i.e. n=3
 $
  {T_3} = a{r^{3 - 1}} \\
  {T_3} = {G_2} \\
  {G_2} = a{r^2} \\
  r = \sqrt[3]{{\dfrac{b}{a}}} \\
  {G_2} = a \times {\left( {\sqrt[3]{{\dfrac{b}{a}}}} \right)^2} \\
  $
Substitute the values of $ {G_1},{G_2} $ in $ \dfrac{{{G_1}^2}}{{{G_2}}} + \dfrac{{{G_2}^2}}{{{G_1}}} $
 $
   = \dfrac{{{G_1}^2}}{{{G_2}}} + \dfrac{{{G_2}^2}}{{{G_1}}} \\
   = \dfrac{{{{\left( {a \times \sqrt[3]{{\dfrac{b}{a}}}} \right)}^2}}}{{a \times {{\left( {\sqrt[3]{{\dfrac{b}{a}}}} \right)}^2}}} + \dfrac{{{{\left( {a \times {{\left( {\sqrt[3]{{\dfrac{b}{a}}}} \right)}^2}} \right)}^2}}}{{a \times \sqrt[3]{{\dfrac{b}{a}}}}} \\
   = \dfrac{{{a^2} \times {{\left( {\sqrt[3]{{\dfrac{b}{a}}}} \right)}^2}}}{{a \times {{\left( {\sqrt[3]{{\dfrac{b}{a}}}} \right)}^2}}} + \dfrac{{{a^2} \times {{\left( {\sqrt[3]{{\dfrac{b}{a}}}} \right)}^4}}}{{a \times \sqrt[3]{{\dfrac{b}{a}}}}} \\
   = a + \left( {a \times {{\left( {\sqrt[3]{{\dfrac{b}{a}}}} \right)}^3}} \right) \\
   = a + \left( {a \times \dfrac{b}{a}} \right) \\
   = a + b \\
   = 2A(\because eq(1)) \\
  $
Therefore, the value of $ \dfrac{{{G_1}^2}}{{{G_2}}} + \dfrac{{{G_2}^2}}{{{G_1}}} $ is 2A.
So, the correct answer is “Option A”.

Note: A geometric progression (G.P.) is a sequence in which each term is derived by multiplying the preceding term by a fixed number called the common ratio (r).Arithmetic mean is the sum of the terms divided by the no. of terms. Geometric mean is the square root of the product of the terms.