
If H is harmonic mean between P and Q. Then the value of \[\dfrac{H}{P} + \dfrac{H}{Q}\] is
A) \[2\]
B) $\dfrac{PQ}{(P+Q)}$
C) \[\dfrac{1}{2}\]
D) $\dfrac{(P+Q)}{PQ}$
Answer
493.8k+ views
Hint: Mean is the most commonly used measure of central tendency. Harmonic mean (HM) is defined as the reciprocal of the arithmetic mean of the reciprocal of the observations. We will use the harmonic mean formula, \[h = \dfrac{{2ab}}{{a + b}}\] where h is the harmonic mean of a and b. HM is used in averaging the ratios. And then, we will substitute the value to get the final output.
Complete step by step answer:
Here, we are given that,
$H$ is the harmonic mean between P and Q.
We will use the harmonic formula as below,
\[H = \dfrac{{2PQ}}{{P + Q}}\] -------- (i)
First, we need to find the value of \[\dfrac{H}{P}\].
So, we will divide the denominator of equation (i) by P, we get
\[ \Rightarrow \dfrac{H}{P} = \dfrac{{2PQ}}{{P(P + Q)}}\]
Simplify the above expression we will get,
\[ \Rightarrow \dfrac{H}{P} = \dfrac{{2Q}}{{P + Q}}\]
Second, we need to find the value of \[\dfrac{H}{Q}\].
So, we will divide the denominator of equation (i) by Q, we get
\[ \Rightarrow \dfrac{H}{Q} = \dfrac{{2PQ}}{{Q(P + Q)}}\]
Simplify the above expression we will get,
\[ \Rightarrow \dfrac{H}{Q} = \dfrac{{2P}}{{P + Q}}\]
Now, we need to find the value of
\[\dfrac{H}{P} + \dfrac{H}{Q}\]
Substituting the values in the above expression, we will get,
\[ = \dfrac{{2Q}}{{P + Q}} + \dfrac{{2P}}{{P + Q}}\]
Take the LCM P+Q, we will get,
\[ = \dfrac{{2Q + 2P}}{{P + Q}}\]
Taking number 2 common in the numerator, we will get,
\[ = \dfrac{{2(Q + P)}}{{P + Q}}\]
Rearranging the above expression, we will get,
\[ = \dfrac{{2(P + Q)}}{{P + Q}}\]
\[ = 2\]
Hence, if $H$ is the Harmonic mean between P and Q, then the value of \[\dfrac{H}{P} + \dfrac{H}{Q}\]\[ = 2\].
Note:
Alternative approach:
Since, H is the harmonic mean between P and Q, then the using the formula of harmonic mean, we will get,
\[H = \dfrac{{2PQ}}{{P + Q}}\]
\[ \Rightarrow \dfrac{1}{H} = \dfrac{{P + Q}}{{2PQ}}\]
\[ \Rightarrow \dfrac{2}{H} = \dfrac{{P + Q}}{{PQ}}\] ----- (ii)
Now, we need to find the value of
\[\dfrac{H}{P} + \dfrac{H}{Q}\]
Taking H common here, we will get,
\[ = H(\dfrac{1}{P} + \dfrac{1}{Q})\]
Taking the LCM = PQ we will get,
\[ = H(\dfrac{{Q + P}}{{PQ}})\]
Rearrange the above expression, we will get,
\[ = H(\dfrac{{P + Q}}{{PQ}})\]
Substituting the value taken from (ii), we will get,
\[ = H(\dfrac{2}{H})\]
\[ = 2\]
$\bullet $ The number of observations provides information on the total number of values that are contained in the Dataset. The formula for mean is (Sum of all the observations/Total number of observations). The most important measures of central tendencies are mean, median, mode and range. The different types of means are Arithmetic mean (AM), GM and Harmonic mean (HM). The relation between all three is $GM^2 = AM \times HM$. GM is applied only to positive values, it means that there will be no zero and negative values. The difference between the arithmetic mean and geometric mean is that, we add numbers in arithmetic mean whereas we calculate the product of numbers in geometric mean.
Complete step by step answer:
Here, we are given that,
$H$ is the harmonic mean between P and Q.
We will use the harmonic formula as below,
\[H = \dfrac{{2PQ}}{{P + Q}}\] -------- (i)
First, we need to find the value of \[\dfrac{H}{P}\].
So, we will divide the denominator of equation (i) by P, we get
\[ \Rightarrow \dfrac{H}{P} = \dfrac{{2PQ}}{{P(P + Q)}}\]
Simplify the above expression we will get,
\[ \Rightarrow \dfrac{H}{P} = \dfrac{{2Q}}{{P + Q}}\]
Second, we need to find the value of \[\dfrac{H}{Q}\].
So, we will divide the denominator of equation (i) by Q, we get
\[ \Rightarrow \dfrac{H}{Q} = \dfrac{{2PQ}}{{Q(P + Q)}}\]
Simplify the above expression we will get,
\[ \Rightarrow \dfrac{H}{Q} = \dfrac{{2P}}{{P + Q}}\]
Now, we need to find the value of
\[\dfrac{H}{P} + \dfrac{H}{Q}\]
Substituting the values in the above expression, we will get,
\[ = \dfrac{{2Q}}{{P + Q}} + \dfrac{{2P}}{{P + Q}}\]
Take the LCM P+Q, we will get,
\[ = \dfrac{{2Q + 2P}}{{P + Q}}\]
Taking number 2 common in the numerator, we will get,
\[ = \dfrac{{2(Q + P)}}{{P + Q}}\]
Rearranging the above expression, we will get,
\[ = \dfrac{{2(P + Q)}}{{P + Q}}\]
\[ = 2\]
Hence, if $H$ is the Harmonic mean between P and Q, then the value of \[\dfrac{H}{P} + \dfrac{H}{Q}\]\[ = 2\].
Note:
Alternative approach:
Since, H is the harmonic mean between P and Q, then the using the formula of harmonic mean, we will get,
\[H = \dfrac{{2PQ}}{{P + Q}}\]
\[ \Rightarrow \dfrac{1}{H} = \dfrac{{P + Q}}{{2PQ}}\]
\[ \Rightarrow \dfrac{2}{H} = \dfrac{{P + Q}}{{PQ}}\] ----- (ii)
Now, we need to find the value of
\[\dfrac{H}{P} + \dfrac{H}{Q}\]
Taking H common here, we will get,
\[ = H(\dfrac{1}{P} + \dfrac{1}{Q})\]
Taking the LCM = PQ we will get,
\[ = H(\dfrac{{Q + P}}{{PQ}})\]
Rearrange the above expression, we will get,
\[ = H(\dfrac{{P + Q}}{{PQ}})\]
Substituting the value taken from (ii), we will get,
\[ = H(\dfrac{2}{H})\]
\[ = 2\]
$\bullet $ The number of observations provides information on the total number of values that are contained in the Dataset. The formula for mean is (Sum of all the observations/Total number of observations). The most important measures of central tendencies are mean, median, mode and range. The different types of means are Arithmetic mean (AM), GM and Harmonic mean (HM). The relation between all three is $GM^2 = AM \times HM$. GM is applied only to positive values, it means that there will be no zero and negative values. The difference between the arithmetic mean and geometric mean is that, we add numbers in arithmetic mean whereas we calculate the product of numbers in geometric mean.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

