Question

Let $ {a_1},{a_2}......... $ be positive real numbers in geometric progression. For each $ n $ , let $ {A_n},{G_n},{H_n} $ be respectively, the arithmetic mean, geometric mean & harmonic mean of $ {a_1},{a_2}.........{a_n} $ . Find the expression for the geometric mean of $ {G_1},{G_2},{G_3}........{G_n} $ in terms of $ {A_1},{A_2},{A_3}.........{A_n},{H_1},{H_2},{H_3}..........{H_n} $

Answer 1

Hint: From the question student should understand that this sum is an application of formulae related to Arithmetic Mean , Geometric Mean , Harmonic Mean. First step towards solving this sum is noting down the formulae for sum upto $ n $ terms . Bring it in the simplest possible form in the next step. After this the student should remove the common terms and bring the relation between these means.

Complete step-by-step answer:
In order to solve the numerical first step is to list down the formulae for Arithmetic Mean , Geometric Mean & Harmonic Mean.
 $ {G_k} = {({a_1} \times {a_2} \times {a_3}.........{a_k})^{1/k}}..............(1) $
Where $ k $ is the last term of the expression.
We can simplify equation $ 1 $ as below
 $ {G_k} = {({a_1}r)^{\dfrac{{k - 1}}{2}}}..............(2) $
Following is the formula for Arithmetic progression upto $ k $ terms.
 $ {A_k} = \dfrac{{{a_1} + {a_2} + ......{a_k}}}{k}..........(3) $
 $ {A_k} = \dfrac{{{a_1}(1 + r + .......{r^{k - 1}})}}{k}..........(4) $
 $ {A_k} = \dfrac{{{a_1}({r^k} - 1)}}{{(r - 1)k}}..........(5) $
Noting down the formula for Harmonic Progression upto $ k $ terms.
 $ {H_k} = \dfrac{k}{{\dfrac{1}{{{a_1}}} + \dfrac{1}{{{a_2}}} + \dfrac{1}{{{a_3}}} + .....\dfrac{1}{{{a_k}}}}}..........(6) $
 $ {H_k} = \dfrac{{{a_1}k}}{{1 + \dfrac{1}{r} + ....... + \dfrac{1}{{{r^{k - 1}}}}}}..........(7) $
 $ {H_k} = \dfrac{{{a_1}k(r - 1) \times {r^{k - 1}}}}{{{r^{k - 1}}}}..........(8) $
From Equations $ 2,5,8 $ ,we get the following relation between $ {G_k},{H_k},{A_k} $
 $ {G_k} = {({A_k}{H_k})^{\dfrac{1}{2}}} $
Considering there are infinite number of terms , equation will transform as follows
 $ {\prod\limits_{k = 1}^n G _k} = \prod\limits_{k = 1}^n {{{({A_k}{H_k})}^{\dfrac{1}{2}}}} ................(9) $
Thus expanding RHS of equation $ 9 $ we get following relation
\[{\prod\limits_{k = 1}^n G _k} = {({A_1}{A_2}.......{A_n} \times {H_1}{H_2}........{H_n})^{\dfrac{1}{{2n}}}}\]
Thus the relation of geometric mean in terms of arithmetic mean and Harmonic mean is
\[{\prod\limits_{k = 1}^n G _k} = {({A_1}{A_2}.......{A_n} \times {H_1}{H_2}........{H_n})^{\dfrac{1}{{2n}}}}\]
So, the correct answer is “\[{\prod\limits_{k = 1}^n G _k} = {({A_1}{A_2}.......{A_n} \times {H_1}{H_2}........{H_n})^{\dfrac{1}{{2n}}}}\]
”.

Note: Though this sum looks extremely complicated and difficult to solve, it is easy if the approach is correct. Students are advised to memorize the formula for Arithmetic Mean , Geometric Mean , Harmonic mean for sum upto $ n $ terms. The sum from this chapter should be picked up last if it is of similar type. This is because if the approach is wrong for the sum , it will lead to complete waste of time. This sum is important for Students who are good with application and like to take up challenging numericals.