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Question

Question: Insert two harmonic means between the numbers 5 and 11....

Insert two harmonic means between the numbers 5 and 11.

Explanation

Solution

Hint: A sequence is said to be in harmonic progression if the sequence generated by taking the reciprocal of each term is an arithmetic progression. If we are given two numbers a and b and we are required to insert two numbers H and H’ between a and b, then H is equal to 3aba+2b\dfrac{3ab}{a+2b} and H’ is equal to 3ab2a+b\dfrac{3ab}{2a+b}. Using this formula, we can solve this question.

Complete step-by-step answer:

Before proceeding with the question, we must know the formula that will be required to solve this question.
In sequences and series, we call a sequence harmonic progression when the sequence generated by taking the reciprocal of each term is an arithmetic progression.
In this question, we are required to insert two harmonic means between 5 and 11. Let us denote these two harmonic means as H and H’.
If we have two numbers a and b, then the two harmonic means between these two numbers are given by the formula,
H = 3aba+2b\dfrac{3ab}{a+2b} and H’ = 3ab2a+b\dfrac{3ab}{2a+b} . . . . . . . . . . . (1)
In the question, we are given two numbers 5 and 11. So, substituting a = 5 and b = 11 in equation (1), we get,
H = 3(5)(11)5+2.11\dfrac{3\left( 5 \right)\left( 11 \right)}{5+2.11} and H’ = 3(5)(11)2.5+(11)\dfrac{3\left( 5 \right)\left( 11 \right)}{2.5+\left( 11 \right)}
\Rightarrow H = 3(5)(11)27\dfrac{3\left( 5 \right)\left( 11 \right)}{27} and H’ = 3(5)(11)21\dfrac{3\left( 5 \right)\left( 11 \right)}{21}
\Rightarrow H = 559\dfrac{55}{9} and H’ = 557\dfrac{55}{7}
So, the two harmonic means are H = 559\dfrac{55}{9} and H’ = 557\dfrac{55}{7}.

Note: One can also find the answer for this question if he/she cannot remember the formula for finding the harmonic mean. In that case, we will assume two numbers H and H’ between a and b and we will assume that sequence a, H, H’, b is in harmonic progression. Then we will take the reciprocal of each term which will convert the sequence in an arithmetic progression. Since the common difference of the arithmetic progression is the same through the sequence, we will find the common difference by subtracting pairs of consecutive terms and then we will equate them. After solving these equations, we can find H and H’.