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If one A.M A and two GM p and q are to be inserted between two given numbers, then $\dfrac{{{p}^{2}}}{q}+\dfrac{{{q}^{2}}}{p}=?$

Last updated date: 26th Feb 2024
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Assume a and b be the two numbers. Let p and q be the G.M. and A.M. respectively. Use the formula for G.M. and A.M. and calculate the values of p and q. Use these values of p and q to calculate the required value.

“Complete step-by-step answer:”
Let a and b be the two numbers between which two G.M.’s p and q and one A.M. A are to be inserted. Let us first insert p and q between a and b so that they form a geometric series
$a\ \ \ p\ \ \ q\ \ \ b$
Now the common ratio r between each number must be the same. Therefore, r of this geometric series can be given as
Also, since $\dfrac{p}{a}=\dfrac{q}{p}=\dfrac{b}{q}=r$ (common ratio between consecutive terms),
we get
$a=\dfrac{{{p}^{2}}}{q}\ \ \ ,\ \ \ \ b=\dfrac{{{q}^{2}}}{p}$
Now, arithmetic mean A between a and b can be given as
Putting values of a and b we get
  & A=\dfrac{1}{2}\left( \dfrac{{{p}^{2}}}{q}+\dfrac{{{q}^{2}}}{p} \right) \\
 & \therefore \dfrac{{{p}^{2}}}{q}+\dfrac{{{q}^{2}}}{p}=2A \\

Note: The problems related to arithmetic and geometric mean are based on a simple principle that adding any number of A.M’s and G.M’s between two terms will keep the terms “equidistant” from each other. If A.M 's are added then the “arithmetic distance” or difference between two consecutive terms must be constant while adding G.M.'s one should keep “geometric distance” or ratio between consecutive terms must be constant. Majority of the problems in Arithmetic and Geometric progression can be solved using the basic principles. The concept of arithmetic and geometric means is not limited to just two numbers. The arithmetic mean of n numbers can also be calculated by using formula
Geometric mean of n numbers can also be given by the formula
We should be very careful though, if we had to insert one G.M. between a and b then $\sqrt{ab}$ would be that term but since we had to insert two G.M.’s, we cannot use this term.