Question

Find the correct option
If $a,{a_1},{a_{2,}}............................,{a_{2n - 1}},{a_{2n}},b$ are in Arithmetic Progression and $a,{g_1},{g_{2,}}............................,{g_{2n - 1}},{g_{2n}},b$ are in Geometric Progression and $h$ is the harmonic mean of $a$ and $b$, then, $\dfrac{{{a_1} + {a_{2n}}}}{{{g_1}{g_{2n}}}} + \dfrac{{{a_2} + {a_{2n - 1}}}}{{{g_2}{g_{2n - 1}}}} + ...................... + \dfrac{{{a_n} + {a_{n + 1}}}}{{{g_n}{g_{n + 1}}}}$ is equal to
A. $2nh$
B. $\dfrac{n}{h}$
C. $nh$
D. $\dfrac{{2n}}{h}$

Answer 1

Hint: Using the properties of arithmetic means in an Arithmetic Progression and geometric means in a Geometric Progression between two numbers, the value of the given expression will be calculated in terms of the harmonic mean between the two given numbers to choose the correct option.

Formula Used:
The following formulas regarding arithmetic means and geometric means between two numbers $a$ and $b$ have been used to solve this problem:
1.If $a,{a_1},{a_{2,}}............................,{a_{2n - 1}},{a_{2n}},b$ are in Arithmetic Progression,
then ${a_1},{a_{2,}}............................,{a_{2n - 1}},{a_{2n}}$ are the arithmetic means between $a$ and $b$ .
and ${a_1} + {a_{2n}} = {a_2} + {a_{2n - 1}} = ....................... = {a_n} + {a_{n + 1}} = a + b$
2.If $a,{g_1},{g_{2,}}............................,{g_{2n - 1}},{g_{2n}},b$ are in Geometric Progression,
then, ${g_1},{g_{2,}}............................,{g_{2n - 1}},{g_{2n}}$ are the geometric meanings between $a$ and $b$ .
and ${g_1}{g_{2n}} = {g_2}{g_{2n - 1}} = ........................ = {g_n}{g_{n + 1}} = ab$

Complete step by step solution:
 We have been given that $a,{a_1},{a_{2,}}............................,{a_{2n - 1}},{a_{2n}},b$ are in Arithmetic Progression
and $a,{g_1},{g_{2,}}............................,{g_{2n - 1}},{g_{2n}},b$ are in Geometric Progression and $h$ is the harmonic mean of $a$ and $b$ .
We have to find out the value of the expression $\dfrac{{{a_1} + {a_{2n}}}}{{{g_1}{g_{2n}}}} + \dfrac{{{a_2} + {a_{2n - 1}}}}{{{g_2}{g_{2n - 1}}}} + ...................... + \dfrac{{{a_n} + {a_{n + 1}}}}{{{g_n}{g_{n + 1}}}}$ in terms of the harmonic mean $h$ .
Since, ${a_1},{a_{2,}}............................,{a_{2n - 1}},{a_{2n}}$ are the arithmetic means between $a$ and $b$
then, ${a_1} + {a_{2n}} = {a_2} + {a_{2n - 1}} = ....................... = {a_n} + {a_{n + 1}} = a + b$ ……………………….equation (1)
Since, ${g_1},{g_{2,}}............................,{g_{2n - 1}},{g_{2n}}$ are the geometric means between $a$ and $b$
then, ${g_1}{g_{2n}} = {g_2}{g_{2n - 1}} = ........................ = {g_n}{g_{n + 1}} = ab$ ……………………….equation (2)
It is given that $h$ is the harmonic mean of $a$ and $b$ .
We know that if two numbers are in harmonic progression, then their reciprocals are in arithmetic progression.
As $a$ and $b$ are in harmonic progression, then, $\dfrac{1}{a}$ and $\dfrac{1}{b}$ will be in arithmetic progression.
The arithmetic mean of $\dfrac{1}{a}$ and $\dfrac{1}{b} = \dfrac{{\dfrac{1}{a} + \dfrac{1}{b}}}{2} = \dfrac{{\dfrac{{a + b}}{{ab}}}}{2} = \dfrac{{a + b}}{{2ab}}$ .
Now, $\dfrac{1}{a},\dfrac{{a + b}}{{2ab}},\dfrac{1}{b}$ are in arithmetic progression
So, $a,\dfrac{{2ab}}{{a + b}},b$ are in harmonic progression and $\dfrac{{a + b}}{{2ab}}$ is the harmonic mean of $a$ and $b$ .
Thus, $h = \dfrac{{2ab}}{{a + b}}$ …………………………………equation (3)
Now, we will simplify the given expression putting the values from equation (1) and (2) as follows:
$\dfrac{{{a_1} + {a_{2n}}}}{{{g_1}{g_{2n}}}} + \dfrac{{{a_2} + {a_{2n - 1}}}}{{{g_2}{g_{2n - 1}}}} + ...................... + \dfrac{{{a_n} + {a_{n + 1}}}}{{{g_n}{g_{n + 1}}}}$
$ = \dfrac{{a + b}}{{ab}} + \dfrac{{a + b}}{{ab}} + ...................... + \dfrac{{a + b}}{{ab}}$
$ = \dfrac{{n(a + b)}}{{ab}}$ (Since $\dfrac{{a + b}}{{ab}}$ is added $n$ times)
$ = 2n \times \dfrac{{(a + b)}}{{2ab}}$ (Multiplying numerator and denominator by $2$)
Further simplifying the above and considering the value of $h = \dfrac{{2ab}}{{a + b}}$ from equation (3), we have,
$\dfrac{{{a_1} + {a_{2n}}}}{{{g_1}{g_{2n}}}} + \dfrac{{{a_2} + {a_{2n - 1}}}}{{{g_2}{g_{2n - 1}}}} + ...................... + \dfrac{{{a_n} + {a_{n + 1}}}}{{{g_n}{g_{n + 1}}}} = 2n \times \dfrac{1}{{\dfrac{{2ab}}{{a + b}}}} = \dfrac{{2n}}{h}$
So, the value of the expression is equal to $\dfrac{{2n}}{h}$ .

Option ‘D’ is correct

Note: The common difference in an A.P. is found by subtracting each term from its succeeding term, where the common ratio in a G.P. is found by dividing each term by its preceding term. If some numbers are in A.P., then their reciprocals will be in H.P.