Question

If the H.M of ‘a’ and ‘b’ is $\dfrac{{kab}}{{a + b}}$ then find the value of ${k^3} - {k^2} + 2k + 1$ .

Answer 1

- Hint: Here we will use the formula of harmonic mean of two given number $a$ and $b$ then we will equate them with $\dfrac{{kab}}{{a + b}}$ and find the value of $k$. After that we will put the value of $k$ in the given question.

Complete step-by-step solution :
Since we know that the Harmonic Mean ‘H’ between two numbers $a$ and $b$ can be given as $H = \dfrac{{2ab}}{{a + b}}$ .
But it is given that $H = \dfrac{{kab}}{{a + b}}$
Thus $\dfrac{{2ab}}{{a + b}}$ $= \dfrac{{kab}}{{a + b}}$
Therefore we get k=2
On putting k=2 in given question ${k^3} - {k^2} + 2k + 1$ we get
${2^3} - {2^2} + 2 \times 2 + 1$
=$8 - 4 + 4 + 1 = 9$ (Ans.)

Hence the required answer is $9.$

Note:
Harmonic Mean: If three terms $a, b$ and $c$ are in Harmonic Progression
Then$\dfrac{1}{a}$, $\dfrac{1}{b}$and $\dfrac{1}{c}$ form an Arithmetic Progression.
Therefore, harmonic mean formula-
$\dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}$
So, the harmonic mean $b = \dfrac{{2ac}}{{a + c}}$
Harmonic Progression A series of terms is known as a HP series when their reciprocals are in arithmetic progression.
Example: $\dfrac{1}{a},\dfrac{1}{{a + d}},\dfrac{1}{{a + 2d}}$ and so on are in HP because a, a + d, a + 2d are in AP.

The nth term of a HP series is ${T_n}$ =$\dfrac{1}{a+(n-1)d}$.

In order to solve a problem on Harmonic Progression, one should make the corresponding AP series and then solve the problem.

$n^th$ term of H.P. = $\dfrac{1}{\text{nth term of corresponding A.P.}}$

If three terms a, b and c are in HP then $b = \dfrac{{2ac}}{{a + c}}$
It is not possible for a harmonic progression (other than the trivial case where $a = 1$ and $k = 0$) to sum to an integer. The reason is that, necessarily, at least one denominator of the progression will be divisible by a prime number that does not divide any other denominator.