Courses
Courses for Kids
Free study material
Offline Centres
More
Store

The harmonic mean of the reciprocal of even numbers 12 to 190 is. \begin{align} & a)\dfrac{1}{101} \\ & b)\dfrac{2}{111} \\ & c)\dfrac{1}{111} \\ & d)\dfrac{2}{101} \\ \end{align}

Last updated date: 17th Jun 2024
Total views: 372.9k
Views today: 9.72k
Verified
372.9k+ views
Hint: Now to find the harmonic mean of the given terms we will first check the corresponding AP. Now we know that the arithmetic mean of an AP is given by the formula $\dfrac{a+l}{2}$ where a is the first term and l is the last term. Hence we will find the arithmetic mean and then take its reciprocal to find harmonic mean.

Complete step by step solution:
Now let us first understand three types of sequence.
There are 3 major types of sequence which are Arithmetic progression, Geometric progression and Harmonic progression.
Now Arithmetic progression is the progression in which the difference between two consecutive terms is equal. Hence the sequence looks like $a,a+d,a+2d,a+3d,....a+\left( n-1 \right)d$ .
Now similarly Geometric progression is a sequence in which the ratio of two consecutive terms are equal. Hence the sequence looks like $a,ar,a{{r}^{2}},a{{r}^{3}},...a{{r}^{n}}$ .
Now let us understand Harmonic sequences.
Harmonic sequences are sequences which are reciprocal of AP. Hence if $a,a+d,a+2d,a+3d,....a+\left( n-1 \right)d$ is an AP then the Harmonic sequence is given by $\dfrac{1}{a},\dfrac{1}{a+d},\dfrac{1}{a+2d},.....\dfrac{1}{a+\left( n-1 \right)d}$
Now the Mean of two numbers is the middle term lying between two terms in the sequence.
Now consider the given sequence $\dfrac{1}{12},\dfrac{1}{14},...\dfrac{1}{190}$
Now we want to find the harmonic mean of the sequence.
Let us check the corresponding AP to the sequence,
Hence have 12, 14, …, 190.
Now we know that the arithmetic mean between a and l is given by $\dfrac{a+l}{2}$
Hence the arithmetic mean between 12 and 190 is $\dfrac{12+190}{2}=\dfrac{202}{2}=101$
Now arithmetic mean is 101. Hence Harmonic mean will be $\dfrac{1}{101}$ .
Hence the harmonic mean of $\dfrac{1}{12}$ and $\dfrac{1}{190}$ is $\dfrac{1}{101}$

So, the correct answer is “Option a”.

Note: Now note that the Harmonic mean of two numbers a and b can also be given by the formula $\dfrac{2ab}{a+b}$ . Now since a and b are in HP then $\dfrac{1}{a}$ and $\dfrac{1}{b}$ are in AP. Hence we can say that the harmonic mean is $\dfrac{2}{\dfrac{1}{a}+\dfrac{1}{b}}$ . Hence on further simplification we get the formula as $\dfrac{2ab}{a+b}$ .