Question

If $9$ AM’s and HM’s are inserted between $2$ and $3$ and if the harmonic mean H is corresponding to arithmetic mean A, then $A_n + \dfrac{6}{H_n}$ is equal to
$
  1)1 \\
  2)3 \\
  3)5 \\
  4)6 \\
 $

Answer 1

Hint: Here we will use the standard series for the arithmetic and the harmonic expressions and then will use the standard formula for the nth terms of the series and then will take its summation and simplify for the resultant required value.

Complete step-by-step answer:
Let us assume that ${A_1},{A_2},.....{A_9}$ are the AM’s and
Also assume that, ${H_1},{H_2},.....{H_9}$ are the HM’s between a and b.
Given that - $9$ AM’s and HM’s are inserted between $2$ and $3$
Therefore, the total number of terms $ = 9 + 2 = 11$
Consider the Arithmetic progression to be –
$2,{A_1},{A_2},.....{A_9},3$
By using the formula for the nth term in the arithmetic progression –
${a_n} = a + (n - 1)d$
Now for the eleventh term will be –
${a_{11}} = a + (11 - 1)d$
Place the known values in the above expression –
$3 = 2 + (10)d$
Make the required term the subject. When you move any term from one side to the opposite side then the sign of the terms also changes. Positive term becomes negative and vice-versa.
$3 - 2 = (10)d$
Simplify the above expression –
$1 = (10)d$
Term multiplicative on one side if moved to the opposite side then it goes to the denominator.
$d = \dfrac{1}{{10}}$
Now, the terms in the arithmetic progression can be given as –
${A_1} = 2 + d$
${A_n} = 2 + \dfrac{n}{{10}}$ ….. (A)
Let us consider the HP $2,{H_1},{H_2},.....{H_9},3$
Also $\dfrac{1}{2},\dfrac{1}{{{H_1}}},{\dfrac{1}{H}_2},.....\dfrac{1}{{{H_9}}},\dfrac{1}{3}$
Now, ${a_{11}} = a + 100$
Place the known values in the above expression –
$\dfrac{1}{2} + 10D = \dfrac{1}{3}$
Simplify the above expression –
\[
  10D = \dfrac{1}{3} - \dfrac{1}{2} \\
  10D = - \dfrac{1}{6} \\
  D = - \dfrac{1}{{60}} \\
 \]
Now,
$\dfrac{1}{{{H_1}}} = \dfrac{1}{2} - \dfrac{1}{{60}}$
Similarly, nth term can be given as –
$\dfrac{1}{{{H_n}}} = \dfrac{1}{2} - \dfrac{n}{{60}}$
Simplify the above expression –
$\dfrac{6}{{{H_n}}} = 3 - \dfrac{n}{{10}}$ ….. (B)
From the equations (A) and (B) –
${A_n} + \dfrac{6}{{{H_n}}} = 2 + \dfrac{n}{{10}} + 3 - \dfrac{n}{{10}}$
Like terms with the same value and opposite sign cancels each other.
${A_n} + \dfrac{6}{{{H_n}}} = 5$
From the given multiple choices – the third option is the correct answer.
So, the correct answer is “Option 3”.

Note: Harmonic mean can be defined as the type of numerical average and it is calculated by dividing the number of observations by the reciprocal of each of the numbers given in the series while in arithmetic progression the difference between two consecutive terms remains the same. Be careful while simplifying the terms and its sign convention.